Reordering Derivatives of Trace Closures of Regular Languages (Full Version)

TL;DR

This paper introduces syntactic derivative operations for trace closures of regular languages, extending automata constructions to non-regular languages, and identifies conditions under which these automata are finite.

Contribution

It defines new derivative operations for trace closures, analyzes their automata, and introduces the concept of uniform scattering rank to determine finiteness conditions.

Findings

Antimirov and Brzozowski automata can be finite for certain trace closures.

Finite automata can be obtained by quotienting for star-connected expressions.

The refined Antimirov automaton truncation remains finite and accepts trace closures.

Abstract

We provide syntactic derivative-like operations, defined by recursion on regular expressions, in the styles of both Brzozowski and Antimirov, for trace closures of regular languages. Just as the Brzozowski and Antimirov derivative operations for regular languages, these syntactic reordering derivative operations yield deterministic and nondeterministic automata respectively. But trace closures of regular languages are in general not regular, hence these automata cannot generally be finite. Still, as we show, for star-connected expressions, the Antimirov and Brzozowski automata, suitably quotiented, are finite. We also define a refined version of the Antimirov reordering derivative operation where parts-of-derivatives (states of the automaton) are nonempty lists of regular expressions rather than single regular expressions. We define the uniform scattering rank of a language and show that, for a regexp whose language has finite uniform scattering rank, the truncation of the (generally infinite) refined Antimirov automaton, obtained by removing long states, is finite without any quotienting, but still accepts the trace closure. We also show that star-connected languages have finite uniform scattering rank.

Equations

$$\small\begin{array}[]{rcl@{\qquad}rcl}\llbracket a\rrbracket&=_{\mathrm{df}}&\{a\}&\llbracket 1\rrbracket&=_{\mathrm{df}}&\mathbf{1}\\ \llbracket 0\rrbracket&=_{\mathrm{df}}&\emptyset&\llbracket EF\rrbracket&=_{\mathrm{df}}&\llbracket E\rrbracket\cdot\llbracket F\rrbracket\\ \llbracket E+F\rrbracket&=_{\mathrm{df}}&\llbracket E\rrbracket\cup\llbracket F\rrbracket&\llbracket E^{*}\rrbracket&=_{\mathrm{df}}&\mu X.\,\mathbf{1}\cup\llbracket E\rrbracket\cdot X\end{array}$$

$$\begin{array}[]{rclrcl}b{\downarrow}&=_{\mathrm{df}}&\mathsf{ff}&D_{a}b&=_{\mathrm{df}}&\mathsf{if\leavevmode\nobreak\ }a=b\mathsf{\leavevmode\nobreak\ then\leavevmode\nobreak\ }1\mathsf{\leavevmode\nobreak\ else\leavevmode\nobreak\ }0\\ 0{\downarrow}&=_{\mathrm{df}}&\mathsf{ff}&D_{a}0&=_{\mathrm{df}}&0\\ (E+F){\downarrow}&=_{\mathrm{df}}&E{\downarrow}\vee F{\downarrow}&D_{a}(E+F)&=_{\mathrm{df}}&D_{a}E+D_{a}F\\ 1{\downarrow}&=_{\mathrm{df}}&\mathsf{tt}&D_{a}1&=_{\mathrm{df}}&0\\ (EF){\downarrow}&=_{\mathrm{df}}&E{\downarrow}\wedge F{\downarrow}&D_{a}(EF)&=_{\mathrm{df}}&\mathsf{if\leavevmode\nobreak\ }E{\downarrow}\mathsf{\leavevmode\nobreak\ then\leavevmode\nobreak\ }(D_{a}E)F+D_{a}F\mathsf{\leavevmode\nobreak\ else\leavevmode\nobreak\ }(D_{a}E)F\\ (E^{*}){\downarrow}&=_{\mathrm{df}}&\mathsf{tt}&D_{a}(E^{*})&=_{\mathrm{df}}&(D_{a}E)E^{*}\\[8.61108pt] &&&D_{\varepsilon}E&=_{\mathrm{df}}&E\\ &&&D_{ua}E&=_{\mathrm{df}}&D_{a}(D_{u}E)\end{array}$$

$$\small\begin{array}[]{l}a\to(a,1)\quad E+F\to(a,E^{\prime})E\to(a,E^{\prime})\quad E+F\to(a,F^{\prime})F\to(a,F^{\prime})\\[7.74997pt] EF\to(a,E^{\prime}F)E\to(a,E^{\prime})\quad EF\to(a,F^{\prime})\lx@proof@logical@and E{\downarrow}F\to(a,F^{\prime})\quad E^{*}\to(a,E^{\prime}E^{*})E\to(a,E^{\prime})\\[7.74997pt] E\to^{*}(\varepsilon,E)\quad E\to^{*}(ua,E^{\prime\prime})\lx@proof@logical@and E\to^{*}(u,E^{\prime})E^{\prime}\to(a,E^{\prime\prime})\end{array}$$

$$\begin{array}[]{rcl}E^{\to^{*}}&=_{\mathrm{df}}&\{E\}\cup E^{\to^{+}}\\[8.61108pt] a^{\to^{+}}&=_{\mathrm{df}}&\{1\}\\ 0^{\to^{+}}&=_{\mathrm{df}}&\emptyset\\ (E+F)^{\to^{+}}&=_{\mathrm{df}}&E^{\to^{+}}\cup F^{\to^{+}}\\ 1^{\to^{+}}&=_{\mathrm{df}}&\emptyset\\ (EF)^{\to^{+}}&=_{\mathrm{df}}&E^{\to^{+}}\cdot\{F\}\cup F^{\to^{+}}\\ (E^{*})^{\to^{+}}&=_{\mathrm{df}}&E^{\to^{+}}\cdot\{E^{*}\}\end{array}$$

$$\begin{array}[]{rcl}\varepsilon\cdot^{I}v&=_{\mathrm{df}}&\{v\}\\ u\cdot^{I}\varepsilon&=_{\mathrm{df}}&\{u\}\\ au\cdot^{I}bv&=_{\mathrm{df}}&\{a\}\cdot(u\cdot^{I}bv)\cup\{b\mid auIb\}\cdot(au\cdot^{I}v)\\[8.61108pt] \end{array}$$

$$L{\cdot }^I{L}^{\prime }{=}_{\mathrm{df}}\bigcup \{u{\cdot }^{I}v\mid u\in L\wedge v\in L^{\prime }\}$$

$$\begin{array}[]{rcl@{\quad}rcl}\mathbf{1}\cdot^{I}L&=&L&\emptyset\cdot^{I}L&=&\emptyset\\ L\cdot^{I}\mathbf{1}&=&L&(L_{1}\cup L_{2})\cdot^{I}L&=&L_{1}\cdot^{I}L\cup L_{2}\cdot^{I}L\\ (L\cdot^{I}L^{\prime})\cdot^{I}L^{\prime\prime}&=&L\cdot^{I}(L^{\prime}\cdot^{I}L^{\prime\prime})&(L_{1}\mathrel{{\sqcup}{\sqcup}}L_{2})\cdot^{I}(L_{1}^{\prime}\mathrel{{\sqcup}{\sqcup}}L_{2}^{\prime})&\subseteq&(L_{1}\cdot^{I}L_{1}^{\prime})\mathrel{{\sqcup}{\sqcup}}(L_{2}\cdot^{I}L_{2}^{\prime})\end{array}$$

$$\small\begin{array}[]{rcl@{\qquad}rcl}\llbracket a\rrbracket^{I}&=_{\mathrm{df}}&\{a\}&\llbracket 1\rrbracket^{I}&=_{\mathrm{df}}&\mathbf{1}\\ \llbracket 0\rrbracket^{I}&=_{\mathrm{df}}&\emptyset&\llbracket EF\rrbracket^{I}&=_{\mathrm{df}}&\llbracket E\rrbracket^{I}\cdot^{I}\llbracket F\rrbracket^{I}\\ \llbracket E+F\rrbracket^{I}&=_{\mathrm{df}}&\llbracket E\rrbracket^{I}\cup\llbracket F\rrbracket^{I}&\llbracket E^{*}\rrbracket^{I}&=_{\mathrm{df}}&\mu X.\,\mathbf{1}\cup\llbracket E\rrbracket^{I}\cdot^{I}X\end{array}$$

$$\small\begin{array}[]{rcl@{\qquad}rcl}R^{I}_{a}b&=_{\mathrm{df}}&\mathsf{if\leavevmode\nobreak\ }aIb\mathsf{\leavevmode\nobreak\ then\leavevmode\nobreak\ }b\mathsf{\leavevmode\nobreak\ else\leavevmode\nobreak\ }0&D^{I}_{a}b&=_{\mathrm{df}}&\mathsf{if\leavevmode\nobreak\ }a=b\mathsf{\leavevmode\nobreak\ then\leavevmode\nobreak\ }1\mathsf{\leavevmode\nobreak\ else\leavevmode\nobreak\ }0\\ R^{I}_{a}0&=_{\mathrm{df}}&0&D^{I}_{a}0&=_{\mathrm{df}}&0\\ R^{I}_{a}(E+F)&=_{\mathrm{df}}&R^{I}_{a}E+R^{I}_{a}F&D^{I}_{a}(E+F)&=_{\mathrm{df}}&D^{I}_{a}E+D^{I}_{a}F\\ R^{I}_{a}1&=_{\mathrm{df}}&1&D^{I}_{a}1&=_{\mathrm{df}}&0\\ R^{I}_{a}(EF)&=_{\mathrm{df}}&(R^{I}_{a}E)(R^{I}_{a}F)&D^{I}_{a}(EF)&=_{\mathrm{df}}&(D^{I}_{a}E)F+(R^{I}_{a}E)(D^{I}_{a}F)\\ R^{I}_{a}(E^{*})&=_{\mathrm{df}}&(R^{I}_{a}E)^{*}&D^{I}_{a}(E^{*})&=_{\mathrm{df}}&(R^{I}_{a}E)^{*}(D^{I}_{a}E)E^{*}\\[7.74997pt] R^{I}_{\varepsilon}E&=_{\mathrm{df}}&E&D^{I}_{\varepsilon}E&=_{\mathrm{df}}&E\\ R^{I}_{ua}E&=_{\mathrm{df}}&R^{I}_{a}(R^{I}_{u}E)&D^{I}_{ua}E&=_{\mathrm{df}}&D^{I}_{a}(D^{I}_{u}E)\end{array}$$

$$\begin{array}[]{rcl}\small D^{I}_{b}E&=&D^{I}_{b}aa+D^{I}_{b}ab+D^{I}_{b}b\\ &=&((D^{I}_{b}a)a+(R^{I}_{b}a)(D^{I}_{b}a))+((D^{I}_{b}a)b+(R^{I}_{b}a)(D^{I}_{b}b))+D^{I}_{b}b\\ &=&(0a+a0)+(0b+a1)+1\doteq a+1\\[8.61108pt] D^{I}_{b}(E^{*})&=&(R^{I}_{b}E)^{*}(D^{I}_{b}E)E^{*}\\ &=&(aa+a0+0)^{*}((0a+a0)+(0b+a1)+1)E^{*}\doteq(aa)^{*}(a+1)E^{*}\\[8.61108pt] D^{I}_{bb}(E^{*})&\doteq&D^{I}_{b}((aa)^{*}(a+1)E^{*})\doteq(aa)^{*}(a+1)(aa)^{*}(a+1)E^{*}\end{array}$$

$$\small\begin{array}[]{l}a\to^{I}(a,1)\quad E+F\to^{I}(a,E^{\prime})E\to^{I}(a,E^{\prime})\quad E+F\to^{I}(a,F^{\prime})F\to^{I}(a,F^{\prime})\\[7.74997pt] EF\to^{I}(a,E^{\prime}F)E\to^{I}(a,E^{\prime})\quad EF\to^{I}(a,(R^{I}_{a}E)F^{\prime})F\to^{I}(a,F^{\prime})\quad E^{*}\to^{I}(a,(R^{I}_{a}E)^{*}E^{\prime}E^{*})E\to^{I}(a,E^{\prime})\\[7.74997pt] E\to^{I*}(\varepsilon,E)\quad E\to^{I*}(ua,E^{\prime\prime})\lx@proof@logical@and E\to^{I*}(u,E^{\prime})E^{\prime}\to^{I}(a,E^{\prime\prime})\end{array}$$

$$\small\begin{array}[]{l}aa+ab+b\to^{I}(b,a1)ab+b\to^{I}(b,a1)ab\to^{I}(b,a1)b\to^{I}(b,1)\qquad aa+ab+b\to^{I}(b,1)ab+b\to^{I}(b,1)b\to^{I}(b,1)\end{array}$$

$$E^{\prime }\doteq (R_{X_0}^{I}E{)}^{\ast }(R_{X_1}^I{E}_1)(R_{X_1}^{I}E{)}^{\ast }\dots (R_{X_{n-1}}^I{E}_{n-1})(R_{X_{n-1}}^{I}E{)}^{\ast }{E}_n{E}^{\ast }$$

$$\small\begin{array}[]{rcl}a^{\to+}&=_{\mathrm{df}}&\{1\}\\ 0^{\to+}&=_{\mathrm{df}}&\emptyset\\ (E+F)^{\to+}&=_{\mathrm{df}}&E^{\to+}\cup F^{\to+}\\ 1^{\to+}&=_{\mathrm{df}}&\emptyset\\ (EF)^{\to+}&=_{\mathrm{df}}&E^{\to+}\cdot\{F\}\cup\bigcup\{R^{I}_{X}(E^{\to*})\cdot F^{\to+}\mid\emptyset\subset X\subseteq\Sigma\}\\ (E^{*})^{\to+}&=_{\mathrm{df}}&\{(R^{I}_{X_{0}}E)^{*}(R^{I}_{X_{1}}E_{1})\ldots(R^{I}_{X_{n-1}}E_{n-1})(R^{I}_{X_{n-1}}E)^{*}E_{n}E^{*}\mid\\ &&\hskip 14.22636ptn>0,\emptyset\subset X_{i}\subseteq\Sigma,X_{i-1}\supseteq X_{i},E_{i}\in E^{\to+}\}\\[7.74997pt] E^{\to*}&=_{\mathrm{df}}&\{E\}\cup E^{\to+}\end{array}$$

$$\small\begin{array}[]{l}a\to^{I}(a;1,1)\quad E+F\to^{I}(a;E_{l},E_{r})E\to^{I}(a;E_{l},E_{r})\quad E+F\to^{I}(a;F_{l},F_{r})F\to^{I}(a;F_{l},F_{r})\\[7.74997pt] EF\to^{I}(a;E_{l},E_{r}F)E\to^{I}(a;E_{l},E_{r})\quad EF\to^{I}(a;(R^{I}_{a}E)F_{l},F_{r})F\to^{I}(a;F_{l},F_{r})\quad E^{*}\to^{I}(a;(R^{I}_{a}E)^{*}E_{l},E_{r}E^{*})E\to^{I}(a;E_{l},E_{r})\end{array}$$

$$\small\begin{array}[]{l}\Gamma,E,\Delta\Rightarrow^{I}_{N}(a;R^{I}_{a}\Gamma,E_{l},E_{r},\Delta)\lx@proof@logical@and E\to^{I}(a;E_{l},E_{r})|\Gamma,\Delta|<N\quad\Gamma,E,\Delta\Rightarrow^{I}_{N}(a;R^{I}_{a}\Gamma,E_{r},\Delta)\lx@proof@logical@and E\to^{I}(a;E_{l},E_{r})E_{l}{\downarrow}|\Gamma|>0\\[7.74997pt] \Gamma,E,\Delta\Rightarrow^{I}_{N}(a;R^{I}_{a}\Gamma,E_{l},\Delta)\lx@proof@logical@and E\to^{I}(a;E_{l},E_{r})E_{r}{\downarrow}|\Delta|>0\quad\Gamma,E,\Delta\Rightarrow^{I}_{N}(a;R^{I}_{a}\Gamma,\Delta)\lx@proof@logical@and E\to^{I}(a;E_{l},E_{r})E_{l}{\downarrow}E_{r}{\downarrow}|\Gamma|>0|\Delta|>0\end{array}$$

$$\small\begin{array}[]{l}E\to^{I*}_{N}(\varepsilon;E)\quad E\to^{I*}_{N}(ua;\Gamma^{\prime})\lx@proof@logical@and E\to^{I*}_{N}(u;\Gamma)\Gamma\Rightarrow^{I}_{N}(a;\Gamma^{\prime})\par\end{array}$$

$$v_{l}Ia\wedge v_{l}a{v}_r\in [[E]]⟺\exists E_l,E_r.E{\to }^I(a;E_l,E_r)\wedge v_l\in [[E_l]]\wedge v_r\in [[E_r]];$$

$$\begin{array}[]{l}\exists z\in\llbracket E\rrbracket,u_{1},\ldots,u_{n}\in\Sigma^{+}.\,u\sim^{I}u_{1}\ldots u_{n}\wedge u_{1},\ldots,u_{n}\lhd z\rhd v_{0},\ldots,v_{n}\\ \iff\\ \exists E_{0},\ldots,E_{n}.\,E\to^{I*}(u;E_{0},\ldots,E_{n})\wedge\forall j.\ v_{j}\in\llbracket E_{j}\rrbracket.\end{array}$$

$$\begin{array}[]{l}(\forall u^{\prime},u^{\prime\prime}.\,u=u^{\prime}u^{\prime\prime}\Longrightarrow\exists v.\,u^{\prime}\mathrel{{\sim}{\lhd}}_{N}z\rhd v)\\ \Longrightarrow\\ \exists E_{0},\ldots,E_{n}.\,E\to^{I*}_{N}(u;E_{0},\ldots,E_{n})\wedge\forall j.\ v_{j}\in\llbracket E_{j}\rrbracket\\ \quad\textrm{for the unique\leavevmode\nobreak\ }n,v_{0},\ldots,v_{n}\textrm{\leavevmode\nobreak\ such that\leavevmode\nobreak\ }u\mathrel{{\sim}{\lhd}}z\rhd v_{0},\ldots,v_{n}\end{array}$$

$$\begin{array}[]{l}\small E^{*}\to^{I*}_{2}(bb;E_{b}^{*}(a1),1(E_{b}^{*}(a1)),1E^{*})\lx@proof@logical@and E^{*}\to^{I*}_{2}(b;E_{b}^{*}(a1),1E^{*})\lx@proof@logical@and E^{*}\to^{I*}_{2}(\varepsilon;E^{*})E^{*}\Rightarrow^{I}_{2}(b;E_{b}^{*}(a1),1E^{*})\lx@proof@logical@and E^{*}\to^{I}(b;E^{*}_{b}(a1),1E^{*})aa+ab+b\to^{I}(b;a1,1)ab+b\to^{I}(b;a1,1)ab\to^{I}(b;a1,1)b\to^{I}(b;1,1)0<2E_{b}^{*}(a1),1E^{*}\Rightarrow^{I}_{2}(b;E_{b}^{*}(a1),1(E_{b}^{*}(a1)),1E^{*})\lx@proof@logical@and 1E^{*}\to^{I}(b;1(E^{*}_{b}(a1)),1E^{*})E^{*}\to^{I}(b;E^{*}_{b}(a1),1E^{*})aa+ab+b\to^{I}(b;a1,1)ab+b\to^{I}(b;a1,1)ab\to^{I}(b;a1,1)b\to^{I}(b;1,1)1<2\end{array}$$

$$\small\begin{array}[]{rcl@{\hspace*{3cm}}rcl}a^{\rightsquigarrow+}&=_{\mathrm{df}}&\{1\}\hfil\hskip 85.35826pt&(E+F)^{\rightsquigarrow+}&=_{\mathrm{df}}&E^{\rightsquigarrow+}\cup F^{\rightsquigarrow+}\\ 0^{\rightsquigarrow+}&=_{\mathrm{df}}&\emptyset\hfil\hskip 85.35826pt&1^{\rightsquigarrow+}&=_{\mathrm{df}}&\emptyset\\[3.87498pt] (EF)^{\rightsquigarrow+}&=_{\mathrm{df}}&\lx@intercol E^{\rightsquigarrow+}\cup F^{\rightsquigarrow+}\cup E^{\rightsquigarrow+}\cdot\{F\}\cup\{E\}\cdot F^{\rightsquigarrow+}\cup E^{\rightsquigarrow+}\cdot F^{\rightsquigarrow+}\hfil\lx@intercol\\ (E^{*})^{\rightsquigarrow+}&=_{\mathrm{df}}&\lx@intercol E^{\rightsquigarrow+}\cup\{E^{*}\}\cdot E^{\rightsquigarrow+}\cup E^{\rightsquigarrow+}\cdot\{E^{*}\}\cup E^{\rightsquigarrow+}\cdot(\{E^{*}\}\cdot E^{\rightsquigarrow+})\cup\ (E^{\rightsquigarrow+}\cdot\{E^{*}\})\cdot E^{\rightsquigarrow+}\hfil\lx@intercol\end{array}$$

$$\small\begin{array}[]{rcl}\mathbf{R}E&=_{\mathrm{df}}&\{R^{I}_{X}E\mid X\subseteq\Sigma\}\\ E^{\to+}&=_{\mathrm{df}}&\mathbf{R}(E^{\rightsquigarrow+})\\ E^{\to*}&=_{\mathrm{df}}&\{E\}\cup E^{\to+}\end{array}$$

$$E$$

Full text

Reordering Derivatives of Trace Closures of Regular Languages (Full Version)

Hendrik Maarand

  

Tarmo Uustalu

1 Introduction

Traces were introduced to concurrency theory by Mazurkiewicz [15, 16] as an alternative to words. A word can be seen as a linear order that is labelled with letters of the alphabet. Intuitively, the main idea of traces is that the linear order, corresponding to sequentiality, is replaced with a partial order. Sets of words (or word languages) can be used to describe the behaviour of concurrent systems. Similarly, sets of traces (or trace languages) can also be used for this purpose. The difference is that descriptions in terms of traces do not distinguish between different linear extensions (words) of the same partial order (trace)—they are considered equivalent. Different linear extensions of the same partial order can be seen as different observations of the same behaviour.

Given a word language $L$ and a letter $a$, the derivative of $L$ along $a$ is the language consisting of all the words $v$ such that $av$ belongs to $L$. An essential difference between words and traces is that a nonempty word (a linear order) has its first letter as the unique minimal element, but a nonempty trace (a partial order) may have several minimal elements. A trace from a trace language can be derived along any of its minimal letters. Clearly, a minimal letter of a trace need not be the first letter of a word representing this trace.

It is well-known that the derivative of a regular word language along a letter is again regular. Brzozowski [7] showed that a regexp for it can be computed from a regexp for the given language, and Antimirov [3] then further optimized this result. We show that these syntactic derivative operations generalize to trace closures (i.e., closures under equivalence) of regular word languages in the form of syntactic reordering derivative operations.

The syntactic derivative operations for regular word languages provide ways to construct automata from a regexp. The Brzozowski derivative operation is a function on regexps while the Antimirov derivative operation is a relation. Accordingly, they yield deterministic and nondeterministic automata. The set of Brzozowski derivatives of a regexp (modulo appropriate equations) and the set of Antimirov parts-of-derivatives are finite, hence so are the resulting automata. Our generalizations to trace closures of regular languages similarly give deterministic and nondeterministic automata, but these cannot be finite in general. Still, as we show, for a star-connected expression, the Antimirov and Brzozowski automata, suitably quotiented, are finite. We also develop a finer version of the Antimirov reordering derivative, where parts-of-derivatives are nonempty lists of regexps rather than single regexps, and we show that the set of expressions that can appear in these lists for a given initial regexp is finite. We introduce a new notion of uniform scattering rank of a language (a variant of Hashiguchi’s scattering rank [9]) and show that, for a regexp whose language has finite uniform rank, a truncation of the refined reordering Antimirov automaton accepts its trace closure despite the removed states, and is finite, without any quotienting.

This is an extension of the conference paper [14] with proofs of the most important propositions and background material on classical language derivatives and trace closures of regular languages.

2 Preliminaries on Word Languages

An alphabet $\Sigma$ is a finite set (of letters). A word over $\Sigma$ is a finite sequence of letters. The set $\Sigma^{*}$ of all words over $\Sigma$ is the free monoid on $\Sigma$ with the empty word $\varepsilon$ as the unit and concatenation of words (denoted by ${\cdot}$ that can be omitted) as the multiplication. By $\pi_{X}(u)$ we mean the projection of a word $u$ to a subalphabet $X\subseteq\Sigma$, i.e., $\pi_{X}(u)$ discards from $u$ all letters which are not in $X$. We write $|u|$ for the length of a word $u$ and also $|X|$ for the size of a subalphabet $X$. By $|u|_{a}$ we mean $|\pi_{a}(u)|$, i.e., the number of occurrences of $a$ in $u$. By $\Sigma(u)$ we denote the set of letters that appear in $u$.

A (word) language is a subset of $\Sigma^{*}$. The empty word and concatenation of words lift to word languages via $\mathbf{1}=_{\mathrm{df}}\{\varepsilon\}$ and $L\cdot L^{\prime}=_{\mathrm{df}}\{uv\mid u\in L\wedge v\in L^{\prime}\}$.

2.1 Regular Languages

The set $\mathsf{RE}$ of regular expressions (in short, regexps) over $\Sigma$ is given by the grammar $E,F::=a\mid 0\mid E+F\mid 1\mid EF\mid E^{*}$ where $a$ ranges over $\Sigma$.

The word-language semantics of regular expressions is given by a function $\llbracket\_\rrbracket:\mathsf{RE}\to\mathcal{P}\Sigma^{*}$ defined recursively by

[TABLE]

A word language $L$ is said to be regular (or rational) if $L=\llbracket E\rrbracket$ for some regexp $E$. Kleene algebras are defined by an equational theory. It was shown by Kozen [13] that the set $\{\llbracket E\rrbracket\mid E\in\mathsf{RE}\}$ of all regular languages together with the language operations $\emptyset$, $\cup$, $\mathbf{1}$, ${\cdot}$, $(\_)^{*}$ is the free Kleene algebra on $\Sigma$. An important property for us is that $E\doteq F$ iff $\llbracket E\rrbracket=\llbracket F\rrbracket$ where $\doteq$ refers to valid equations in the Kleene algebra theory.

Kleene’s theorem [11] says that a word language is rational iff it is recognizable, i.e., accepted by a finite deterministic automaton (acceptance by a finite nondeterministic automaton is an equivalent condition because of determinizability [20]).

2.2 Derivatives of a Language

A word language $L$ is said to be nullable $L{\downarrow}$, if $\varepsilon\in L$. The derivative (or left quotient)111We use the word ‘derivative’ both for languages and expressions, reserving the word ‘quotient’ for quotients of sets by equivalence relations. of $L$ along a word $u$ is defined by $D_{u}L=_{\mathrm{df}}\{v\mid uv\in L\}$. For any $L$, we have $D_{\varepsilon}L=L$ as well as $D_{uv}L=D_{v}(D_{u}L)$ for any $u,v\in\Sigma^{*}$, i.e., the operation $D:\mathcal{P}\Sigma^{*}\times\Sigma^{*}\to\mathcal{P}\Sigma^{*}$ is a right action of $\Sigma^{*}$ on $\mathcal{P}\Sigma^{*}$. We also have $L=\{\varepsilon\mid L{\downarrow}\}\cup\bigcup\{\{a\}\cdot D_{a}L\mid a\in\Sigma\}$, and for any $u\in\Sigma^{*}$, we have $u\in L$ iff $(D_{u}L){\downarrow}$.

Derivatives of regular languages are regular. A remarkable fact is that they can be computed syntactically, on the level of regular expressions. There are two constructions for this, due to Brzozowski [7] and Antimirov [3]. We review these in the next two subsections. The Brzozowski and Antimirov derivative operations yield deterministic resp. nondeterministic automata accepting the language of a regular expression $E$. The Antimirov automaton is finite. The Brzozowski automaton becomes finite when quotiented by associativity, commutativity and idempotence for $+$. Identified up to the Kleene algebra theory, the states of the Brzozowski automaton correspond to the derivatives of the language $\llbracket E\rrbracket$. Regular languages can be characterized as languages with finitely many derivatives.

2.3 Brzozowski Derivative

Nullability and derivative are semantic notions, defined about languages. However, Brzozowski [7] noticed that for regular languages, one can compute nullability and the derivatives syntactically, on the level of regular expressions.

Definition 2.1**.**

The syntactic nullability and the Brzozowski derivative of a regexp are given by functions ${{\downarrow}}:\mathsf{RE}\to\mathbb{B}$, $D:\mathsf{RE}\times\Sigma\to\mathsf{RE}$ and $D:\mathsf{RE}\times\Sigma^{*}\to RE$ defined recursively by

[TABLE]

Proposition 2.2**.**

For any $E$,

$\llbracket E\rrbracket{\downarrow}=E{\downarrow}$; 2. 2.

for any $a\in\Sigma$, $D_{a}\llbracket E\rrbracket=\llbracket D_{a}E\rrbracket$; 3. 3.

for any $u\in\Sigma^{*}$, $D_{u}\llbracket E\rrbracket=\llbracket D_{u}E\rrbracket$.

Corollary 2.3**.**

For any $E$,

$\llbracket E\rrbracket=\{\varepsilon\mid E{\downarrow}\}\cup\bigcup\{\{a\}\cdot\llbracket D_{a}E\rrbracket\mid a\in\Sigma\}$; 2. 2.

for any $a\in\Sigma$, $v\in\Sigma^{*}$, $av\in\llbracket E\rrbracket$ iff $v\in\llbracket D_{a}E\rrbracket$; 3. 3.

for any $u,v\in\Sigma^{*}$, $uv\in\llbracket E\rrbracket$ iff $v\in\llbracket D_{u}E\rrbracket$; 4. 4.

for any $u\in\Sigma^{*}$, $u\in\llbracket E\rrbracket$ iff $(D_{u}E){\downarrow}$.

The Brzozowski derivative operation gives a method for turning a regular expression into a deterministic automaton. For a regexp $E$, the set of states is $Q^{E}=\{D_{u}E\mid u\in\Sigma^{*}\}$, the initial state is $q^{E}_{0}=E$, the final states are $F^{E}=\{E^{\prime}\in Q^{E}\mid E^{\prime}{\downarrow}\}$ and the transition function $\delta^{E}$ is defined by $D$ restricted to $Q^{E}$.

This automaton is generally not finite, but its quotient by a suitable syntactically defined equivalence relation on the state set is finite, as we will see in the next subsection.

2.4 Antimirov Derivative

Antimirov [3] optimized Brzozowski’s construction essentially constructing a nondeterministic finite automaton (NFA) instead of a DFA, with a smaller number of states and, crucially, without having to identify states up to equations.

Antimirov’s syntactic derivative operation is a multivalued function, in other words, a relation. Antimirov spoke of “partial derivatives” or “linear factors”, we prefer to use the term “parts-of-derivatives”.

Definition 2.4**.**

The Antimirov parts-of-derivatives of a regular expression along a letter or a word are given the relations ${\to}\subseteq\mathsf{RE}\times\Sigma\times\mathsf{RE}$ and ${\to^{*}}\subseteq\mathsf{RE}\times\Sigma^{*}\times\mathsf{RE}$ defined inductively by

[TABLE]

The Antimirov parts-of-derivatives compute the semantic derivative collectively.222If we took languages to be multisets of words (i.e., introduced the notion of a word occurring in a language some number of times ) and adopted the obvious multisets-of-words semantics of regular expressions, the Antimirov parts-of-derivatives would compute regular expressions for a partition of the semantic derivative. In the sets-of-words semantics, however, overlaps are possible, so we do not get a partition.

Proposition 2.5**.**

For any $E$,

for any $a\in\Sigma$, $D_{a}\llbracket E\rrbracket=\bigcup\{\llbracket E^{\prime}\rrbracket\mid E\to(a,E^{\prime})\}$; 2. 2.

for any $u\in\Sigma^{*}$, $D_{u}\llbracket E\rrbracket=\bigcup\{\llbracket E^{\prime}\rrbracket\mid E\to^{*}(u,E^{\prime})\}$.

Corollary 2.6**.**

For any $E$,

$av\in\llbracket E\rrbracket=\{u\mid\exists E^{\prime}.\,E\to(a,E^{\prime})\wedge v\in\llbracket E^{\prime}\rrbracket\}$. 2. 2.

$uv\in\llbracket E\rrbracket=\{u\mid\exists E^{\prime}.\,E\to^{*}(u,E^{\prime})\wedge v\in\llbracket E^{\prime}\rrbracket\}$. 3. 3.

$u\in\llbracket E\rrbracket=\{u\mid\exists E^{\prime}.\,E\to^{*}(u,E^{\prime})\wedge E^{\prime}{\downarrow}\}$.

The parts-of-derivatives of a regexp $E$ induce a nondeterministic automaton. The state set is $Q^{E}=_{\mathrm{df}}\{E^{\prime}\mid\exists u\in\Sigma^{*}.\,E\to^{*}(u,E^{\prime})\}$. The set of initial states is $I^{E}=_{\mathrm{df}}\{E\}$. The set of final states is $F^{E}=_{\mathrm{df}}\{E^{\prime}\in Q^{E}\mid E^{\prime}{\downarrow}\}$. Finally, the transition relation is defined by $E^{\prime}\to^{E}(a,E^{\prime\prime})=_{\mathrm{df}}E^{\prime}\to(a,E^{\prime\prime})$ for $E^{\prime},E^{\prime\prime}\in Q^{E}$.

The state set $Q^{E}$ is shown finite by proving it to be a subset of another set that is straightforwardly seen to be finite.

Definition 2.7**.**

For any $E$, we define a set $E^{\to^{*}}$ of regexps recursively by

[TABLE]

Proposition 2.8**.**

For any $E$,

$E^{\to^{*}}$* is finite, in fact, of cardinality linear in the size of $E$;* 2. 2.

$Q^{E}\subseteq E^{\to^{*}}$.

Proof 2.9**.**

Both parts by induction on $E$.

Corollary 2.10**.**

For any $E$, the Antimirov automaton is finite.

We note that the Antimirov automaton, constructed as above, while canonical, is generally not trim: every state is reachable, but not every state is generally coreachable (i.e., not every state needs to have a path to some final state). A state $E^{\prime}$ is not coreachable if and only if $\llbracket E^{\prime}\rrbracket=\emptyset$. This is the case precisely when $E^{\prime}$ equals [math] in the theory of idempotence of $+$ and the left and right zero laws of [math] wrt. $\cdot$. The Antimirov automaton is trimmed by removing the states that are not coreachable.

Now we can also show that a suitable quotient of the Brzozowski automaton is finite.

For this we prove a syntactic version of Proposition 2.5 relating the Brzozowski derivative and the Antimirov parts-of-derivatives.

Proposition 2.11**.**

For any $E$,

for any $a\in\Sigma$, $D_{a}E\doteq\sum\{E^{\prime}\mid E\to(a,E^{\prime})\}$; 2. 2.

for any $u\in\Sigma^{*}$, $D_{u}E\doteq\sum\{E^{\prime}\mid E\to^{*}(u,E^{\prime})\}$.

(using the semilattice equations for $0,+$, that [math] is left zero, and distributivity of $\cdot$ over $+$ from the right).

Corollary 2.12**.**

For any $E$, the Brzozowski automaton, suitably quotiented, is finite.

Proof 2.13**.**

Just notice that the powerset of a finite set is finite too.

This quotient does not give the minimal deterministic automaton (given by semantic derivatives of $\llbracket E\rrbracket$). The minimal deterministic automaton is obtained from the Brzozowski automaton by quotienting it by the full Kleene algebra theory.

3 Trace Closures of Regular Languages

3.1 Trace Closure of a Word Language

An independence alphabet is an alphabet $\Sigma$ together with an irreflexive and symmetric relation $I\subseteq\Sigma\times\Sigma$ called the independence relation. The complement $D$ of $I$, which is reflexive and symmetric, is called dependence. We extend independence to words by saying that two words $u$ and $v$ are independent, $uIv$, if $aIb$ for all $a,b$ such that $a\in\Sigma(u)$ and $b\in\Sigma(v)$.

Let $\sim^{I}\subseteq\Sigma^{*}\times\Sigma^{*}$ be the least congruence relation on the free monoid $\Sigma^{*}$ such that $aIb$ implies $ab\sim^{I}ba$ for all $a,b\in\Sigma$. If $uIv$, then $uv\sim^{I}vu$.

A (Mazurkiewicz) trace is an equivalence class of words wrt. $\sim^{I}$. The equivalence class of a word $w$ is denoted by $[w]^{I}$.

A word $a_{1}\ldots a_{n}$ where $a_{i}\in\Sigma$ yields a directed node-labelled acyclic graph as follows. Take the vertex set to be $V=_{\mathrm{df}}\{1,\ldots,n\}$ and label vertex $i$ with $a_{i}$. Take the edge set to be $E=_{\mathrm{df}}\{(i,j)\mid i<j\wedge a_{i}Da_{j}\}$. This graph $(V,E)$ for a word $w$ is called the dependence graph of $w$ and is denoted by $\langle w\rangle_{D}$. If $w\sim^{I}z$, then the dependence graphs of $w$ and $z$ are isomorphic, i.e., traces can be identified with dependence graphs up to isomorphism.

The set $\Sigma^{*}/{\sim^{I}}$ of all traces is the free partially commutative monoid on $(\Sigma,I)$. If $I=\emptyset$, then $\Sigma^{*}/{\sim^{I}}\cong\Sigma^{*}$, the set of words, i.e., we recover the free monoid. If $I=\{(a,b)\mid a\neq b\}$, then $\Sigma^{*}/{\sim^{I}}\cong\mathcal{M}_{\mathrm{f}}(\Sigma)$, the set of finite multisets over $\Sigma$, i.e., the free commutative monoid.

A trace language is a subset of $\Sigma^{*}/{\sim^{I}}$. Trace languages are in bijection with word languages that are (trace) closed in the sense that, if $z\in L$ and $w\sim^{I}z$, then also $w\in L$. If $T$ is a trace language, then its flattening $L=_{\mathrm{df}}\bigcup T$ is a closed word language. On the other hand, the trace language corresponding to a closed word language $L$ is $T=_{\mathrm{df}}\{t\in\Sigma^{*}/{\sim^{I}}\mid\exists z\in t.\,z\in L\}=\{t\in\Sigma^{*}/{\sim^{I}}\mid\forall z\in t.\,z\in L\}$.

Given a general (not necessarily closed) word language $L$, we define its (trace) closure $[L]^{I}$ as the least closed word language that contains $L$. Clearly $[L]^{I}=\{w\in\Sigma^{*}\mid\exists z\in L.\,w\sim^{I}z\}$ and also $[L]^{I}=\bigcup\{t\in\Sigma^{*}/{\sim^{I}}\mid\exists z\in t.\,z\in L\}$. For any $L$, we have $[[L]^{I}]^{I}=[L]^{I}$, so $[\_]^{I}$ is a closure operator. Note also that $L$ is closed iff $[L]^{I}=L$.

As seen in Section 2.2, the derivative of a word language is the set of all suffixes for a prefix. We now look at what the prefixes and suffixes of a word as a representative of a trace should be. For a word $vuv^{\prime}$ such that $vIu$, we can consider $u$ to be its prefix, up to reordering, and $vv^{\prime}$ to be the suffix. This is because an equivalent word $uvv^{\prime}$ strictly has $u$ as a prefix and $vv^{\prime}$ as the suffix. Similarly, we may also want to consider $u^{\prime}$ to be a prefix of $vuv^{\prime}$ when $u^{\prime}\sim^{I}u$ since $u^{\prime}vv^{\prime}\sim^{I}uvv^{\prime}\sim^{I}vuv^{\prime}$. Note that if $a$ is such a prefix of $z$, then, by irreflexivity of $I$, this $a$ is the first $a$ of $z$. In general, when $u$ is a prefix of $z$, then the letter occurrences in $u$ uniquely map to letter occurrences in $z$. We scale these ideas to allow $u$ to be scattered in $z$ as $z=v_{0}u_{1}v_{1}\ldots u_{n}v_{n}$ in either the sense that $u=u_{1}\ldots u_{n}$ or $u\sim^{I}u_{1}\ldots u_{n}$. We also define degree-bounded versions of scattering that become relevant in Section 5.

Definition 3.1**.**

*For all $u_{1},\ldots,u_{n}\in\Sigma^{+},v_{0}\in\Sigma^{*},v_{1}\ldots,v_{n-1}\in\Sigma^{+},v_{n}\in\Sigma^{*},z\in\Sigma^{*}$,

$u_{1},\ldots,u_{n}\lhd z\rhd v_{0},\ldots,v_{n}=_{\mathrm{df}}z=v_{0}u_{1}v_{1}\ldots u_{n}v_{n}\wedge\forall i.\,\forall j<i.\,v_{j}Iu_{i}$.*

Definition 3.2**.**

For all $u,v,z\in\Sigma^{*}$,

$u\lhd z\rhd v=_{\mathrm{df}}\exists n\in\mathbb{N},u_{1},\ldots,u_{n},v_{0},\ldots,v_{n}.\,u=u_{1}\ldots u_{n}\wedge v=v_{0}\ldots v_{n}\wedge\\ \hskip 14.22636ptu_{1},\ldots,u_{n}\lhd z\rhd v_{0},\ldots,v_{n}$; 2. 2.

$u\mathrel{{\sim}{\lhd}}z\rhd v=_{\mathrm{df}}\exists u^{\prime}.\,u\sim^{I}u^{\prime}\wedge u^{\prime}\lhd z\rhd v$; 3. 3.

$u\mathrel{{\sim}{\lhd}}z\mathrel{{\rhd}{\sim}}v=_{\mathrm{df}}\exists u^{\prime},v^{\prime}.\,u\sim^{I}u^{\prime}\wedge u^{\prime}\lhd z\rhd v^{\prime}\wedge v^{\prime}\sim^{I}v$.

In all three cases, we talk about $u$ being a prefix and $v$ being a suffix of $z$, up to reordering, or $uv$ being scattered in $z$ with degree $n$.

Lemma 3.3**.**

For any $u,v,z\in\Sigma^{*}$,

$u\lhd z\rhd v\iff\exists!n\in\mathbb{N},u_{1},\ldots,u_{n},v_{0},\ldots,v_{n}.\,u=u_{1}\ldots u_{n}\wedge v=v_{0}\ldots v_{n}\wedge\\ \hskip 14.22636ptu_{1},\ldots,u_{n}\lhd z\rhd v_{0},\ldots,v_{n}$; 2. 2.

$u\mathrel{{\sim}{\lhd}}z\rhd v\iff\exists!u^{\prime}.\,u\sim^{I}u^{\prime}\wedge u^{\prime}\lhd z\rhd v$; 3. 3.

$u\mathrel{{\sim}{\lhd}}z\mathrel{{\rhd}{\sim}}v\iff\exists!u^{\prime},v^{\prime}.\,u\sim^{I}u^{\prime}\wedge u^{\prime}\lhd z\rhd v^{\prime}\wedge v^{\prime}\sim^{I}v$.

Definition 3.4**.**

For all $u,v,z\in\Sigma^{*}$ and $N\in\mathbb{N}$,

$u\lhd_{N}z\rhd v=_{\mathrm{df}}\exists n\leq N,u_{1},\ldots,u_{n},v_{0},\ldots,v_{n}.\,u_{1},\ldots,u_{n}\lhd z\rhd v_{0},\ldots,v_{n}$; 2. 2.

(and $u\mathrel{{\sim}{\lhd}}_{N}z\rhd v$ and $u\mathrel{{\sim}{\lhd}}_{N}z\mathrel{{\rhd}{\sim}}v$ are defined analogously).

Example 3.5**.**

Let $\Sigma=_{\mathrm{df}}\{a,b,c\}$ and $aIb$ and $aIc$. Take $z=_{\mathrm{df}}aabcba$. We have $ab\lhd z\rhd acba$ since $a,b\lhd z\rhd\varepsilon,a,cba$. We can visualize this by underlining the subwords of $u=_{\mathrm{df}}ab$ in $z=\varepsilon\underline{a}a\underline{b}cba$. This scattering is valid because $\varepsilon Ia$, $\varepsilon Ib$ and $aIb$: recall that Def. 3.1 requires all underlined subwords $u_{i}$ to be independent with all non-underlined subwords $v_{i}$ to their left in $z$. Similarly we have $aa,a\lhd z\rhd\varepsilon,bcb,\varepsilon$ because $z=\varepsilon\underline{aa}bcb\underline{a}\varepsilon$, $\varepsilon Iaa$, $\varepsilon Ia$ and $bcbIa$. Note that neither $aa\underline{b}cb\underline{a}\varepsilon$ nor $aabc\underline{ba}\varepsilon$ satisfies the conditions about independence and thus there is no $v$ such that $ba\lhd z\rhd v$. We do have $ba\mathrel{{\sim}{\lhd}}z\rhd acba$ though, since $ba\sim^{I}ab$ and $a,b\lhd z\rhd\varepsilon,a,cba$.

Proposition 3.6**.**

For all $u,v,z\in\Sigma^{*},uv\sim^{I}z\iff u\mathrel{{\sim}{\lhd}}z\mathrel{{\rhd}{\sim}}v$.

3.2 Trace-Closing Semantics of Regular Expressions

We now define a nonstandard word-language semantics of regexps that directly interprets $E$ as the trace closure $[\llbracket E\rrbracket]^{I}$ of its standard regular word-language denotation of $\llbracket E\rrbracket$.

We have $[\{a\}]^{I}=\{a\}$, $[\emptyset]^{I}=\emptyset$, $[L\cup L^{\prime}]^{I}=[L]^{I}\cup[L^{\prime}]^{I}$ and $[\mathbf{1}]^{I}=\mathbf{1}$. But for general $I$, we do not have $[L\cdot L^{\prime}]^{I}=[L]^{I}\cdot[L^{\prime}]^{I}$. For example, for $\Sigma=_{\mathrm{df}}\{a,b\}$ and $aIb$, we have $[\{a\}]^{I}=\{a\}$, $[\{b\}]^{I}=\{b\}$ whereas $[\{ab\}]^{I}=\{ab,ba\}\neq\{ab\}=[\{a\}]^{I}\cdot[\{b\}]^{I}$. Hence we need a different concatenation operation.

Definition 3.7**.**

The $I$-reordering concatenation of words $\cdot^{I}:\Sigma^{*}\times\Sigma^{*}\to\mathcal{P}\Sigma^{*}$ is defined by

[TABLE] 2. 2.

The lifting of $I$-reordering concatenation to languages is defined by

[TABLE]

Note that $\{b\mid auIb\}$ acts as a test: it is either $\emptyset$ or $\{b\}$.

Example 3.8**.**

Let $\Sigma=_{\mathrm{df}}\{a,b\}$ and $aIb$. Then $a\cdot^{I}b=\{ab,ba\}$, $aa\cdot^{I}b=\{aab,aba,baa\}$, $a\cdot^{I}bb=\{abb,bab,bba\}$ and $ab\cdot^{I}ba=\{abba\}$. The last example shows that although $I$-reordering concatenation is defined quite similarly to shuffle, it is different.

Proposition 3.9**.**

For any $u,v,z\in\Sigma^{*}$, $z\in u\cdot^{I}v\iff u\lhd z\rhd v$.

Proposition 3.10**.**

For any languages $L$ and $L^{\prime}$, $[L\cdot L^{\prime}]^{I}=[L]^{I}\cdot^{I}[L^{\prime}]^{I}$.

Evidently, if $I=\emptyset$, then reordering concatenation is just ordinary concatenation: $u\cdot^{\emptyset}v=\{uv\}$. For $I=\Sigma\times\Sigma$, which is forbidden in independence alphabets, as $I$ is required to be irreflexive, it is shuffle: $u\cdot^{\Sigma\times\Sigma}v=u\mathrel{{\sqcup}{\sqcup}}v$. For general $I$, it has properties similar to concatenation. In particular, we have

[TABLE]

but also other equations of the concurrent Kleene algebra theory introduced in [10].

We are ready to introduce the closing semantics of regular expressions.

Definition 3.11**.**

The trace-closing semantics $\llbracket\_\rrbracket^{I}:\mathsf{RE}\to\mathcal{P}\Sigma^{*}$ of regular expressions is defined recursively by

[TABLE]

Compared to the standard semantics of regular expressions, the difference is in the handling of the $EF$ case (and consequently also the $E^{*}$ case) due to the cross-commutation that happens in concatenation of traces and must be accounted for by ${\cdot^{I}}$.

With $I=\emptyset$, we fall back to the standard interpretation of regular expressions: $\llbracket E\rrbracket^{\emptyset}=\llbracket E\rrbracket$. For $I$ a general independence relation, we obtain the desired property that the semantics delivers the trace closure of the language of the regexp.

Proposition 3.12**.**

For any $E$, $\llbracket E\rrbracket^{I}$ is trace closed; moreover, $\llbracket E\rrbracket^{I}=[\llbracket E\rrbracket]^{I}$.

3.3 Properties of Trace Closures of Regular Languages

Trace closures of regular languages are theoretically interesting due to their intricate properties and have therefore been studied in a number of works, e.g., [4, 18, 2, 21, 9, 12]. For a thorough survey, see Ochmański’s handbook chapter [19].

The most important property for us is that the trace closure of a regular language is not necessarily regular.

Proposition 3.13**.**

There exists a regular language $L$ such that $[L]^{I}$ is not regular.

Proof 3.14**.**

Consider $\Sigma=_{\mathrm{df}}\{a,b\}$, $aIb$. Let $L=_{\mathrm{df}}\llbracket(ab)^{*}\rrbracket$. The language $[L]^{I}=\{u\mid|u|_{a}=|u|_{b}\}$ is not regular.

The class of trace closures of regular languages over an independence alphabet behaves quite differently from the class of regular languages over an alphabet. Here are some results demonstrating this.

Theorem 3.15** **(Bertoni et al. [5], Aalbersberg and Welzl [2],

Sakarovitch [21]).

(cf. [19, Thm. 6.2.5]) The class of trace closures of regular languages over $(\Sigma,I)$ is closed under complement iff $I$ is quasi-transitive (i.e., its reflexive closure is transitive).

Theorem 3.16** **(Bertoni et al. [4], Aalbersberg and

Welzl [2] (“if” part); Aalbersberg and Hoogeboom [1]).

(cf. [19, Thm. 6.2.5]) The problem of whether the trace closures of two regular languages over $(\Sigma,I)$ are equal is decidable iff $I$ is quasi-transitive.

Theorem 3.17** (Sakarovitch [22]).**

(cf. [19, Thm. 6.2.7]) The problem of whether the trace closure of the language of a regexp over $(\Sigma,I)$ is regular is decidable iff $I$ is quasi-transitive.

A closed language is regular iff the corresponding trace language is accepted by a finite asynchronous (a.k.a. Zielonka) automaton [24, 25]. In Section 4.4, we will see further characterizations of regular closed languages based on star-connected expressions.

3.4 Rational and Recognizable Languages of Monoids

Trace languages are a special case of languages of monoids.

A subset $T$ of a monoid $M$ is called an $M$-language.

An $M$-language $T$ is called rational if $T=\llbracket E\rrbracket^{M}$ for some regular expression $E$ over $M$. Here $\llbracket\_\rrbracket^{M}:\mathsf{RE}(M)\to\mathcal{P}M$ interprets any element $m$ of $M$ as $\{m\}$, the $0,+$ constructors of regular expressions by $\emptyset$ and $\cup$, the $1,{\cdot}$ constructors as mandated by the monoid structure, and $(\_)^{*}$ as the appropriate least fixpoint.

An $M$-language $T$ is called recognizable if there is a deterministic finite $M$-automaton accepting $T$. An deterministic $M$-automaton is given by a state set $Q$, an initial state $q_{0}\in Q$, a set of final states $F\subseteq Q$, and a right action $\delta$ of $M$ on $Q$. An element $m\in M$ is accepted by the automaton if $\delta_{m}q_{0}\in F$.

Kleene’s celebrated theorem says that, for languages of free monoids on finite sets (i.e., word languages over finite alphabets), rationality and recognizability are equivalent conditions (and we can thus just speak about regularity). For a general monoid, however, the two notions are different.

Theorem 3.18** (Kleene [11]).**

Let $M$ be the free monoid $\Sigma^{*}$ on a finite set $\Sigma$. An $M$-language $T$ is rational iff $T$ is recognizable.

Theorem 3.19** (McKnight [17]).**

Let $M$ be finitely generated. If an $M$-language $T$ is recognizable, then $T$ is rational.

Given a monoid $M$ and a congruence $\equiv$ on $M$, the set $M/{\equiv}$ is a monoid too. We view $M/{\equiv}$-languages as sets of equivalence classes wrt.  $\equiv$.

Proposition 3.20**.**

Given a monoid $M$ and a congruence $\equiv$ on it.

The $M/{\equiv}$-language $\llbracket E\rrbracket^{M/{\equiv}}$ of a regular expression $E$ is expressible via its $M$-language $\llbracket E\rrbracket^{M}$ by $\llbracket E\rrbracket^{M/{\equiv}}=\{t\in M/{\equiv}\mid\exists u\in t.\,u\in\llbracket E\rrbracket^{M}\}$. 2. 2.

A $M/{\equiv}$-language $T$ is recognizable iff its flattening $\bigcup T$ into an $M$-language is recognizable.

For the monoid $\Sigma^{*}/{\sim^{I}}$ of traces, which is the free partially commutative monoid, the classes of rational and recognizable languages are different, the class of rational languages is a proper subclass of that of recognizable languages. In view of Proposition 3.20, a trace language $T$ is rational iff $T=\{t\in\Sigma^{*}/{\sim^{I}}\mid\exists u\in t.\,u\in L\}$ or, equivalently, $\bigcup T=[L]^{I}$ for some regular word language $L$ (in the alternative terminology of Aalbersberg and Welzl [2], such a trace language $T$ is called existentially regular), and recognizable iff $\bigcup T=L$ for some regular word language $L$ (such a trace language is called consistently regular).

The question of when a rational trace language is recognizable is nontrivial. We have just seen that, reformulated in terms of word languages, it becomes: given a regular language $L$, when is its trace closure $[L]^{I}$ regular?

4 Reordering Derivatives

We are now ready to generalize the Brzozowski and Antimirov constructions for trace closures of regular languages. To this end, we switch to what we call reordering derivatives.

4.1 Reordering Derivative of a Language

Let $(\Sigma,I)$ be a fixed independence alphabet. We generalize the concepts of (semantic) nullability and derivative of a language to concepts of reorderable part and reordering derivative.

Definition 4.1**.**

We define the $I$-reorderable part of a language $L$ wrt. a word $u$ by $R^{I}_{u}L=_{\mathrm{df}}\{v\in L\mid vIu\}$ and the $I$-reordering derivative along $u$ by $D^{I}_{u}L=_{\mathrm{df}}\{v\mid\exists z\in L.\,u\mathrel{{\sim}{\lhd}}z\rhd v\}$.

By Prop. 3.9, we can equivalently say that $D^{I}_{u}L=\{v\mid\exists z\in L.\,z\in[u]^{I}\cdot^{I}v\}$. For a single-letter word $a$, we get $D^{I}_{a}L=\{v_{l}v_{r}\mid v_{l}av_{r}\in L\wedge v_{l}Ia\}=\{v\mid\exists z\in L.\,z\in a\cdot^{I}v\}$. That is, we require some reordering of $u$ (resp. $a$) to be a prefix, up to reordering, of some word $z$ in $L$ with $v$ as the corresponding strict suffix. (In other words, for the sake of precision and emphasis, we allow reordering of letters within $u$ and across $u$ and $v$, but not within $v$.)

Example 4.2**.**

Let $\Sigma=_{\mathrm{df}}\{a,b,c\}$ and $aIb$. Take $L=_{\mathrm{df}}\{\varepsilon,a,b,ca,aa,bbb,babca,abbaba\}$. We have $R^{I}_{a}L=R^{I}_{aa}L=\{\varepsilon,b,bbb\}$, $D^{I}_{a}L=\{\varepsilon,a,bbca,bbaba\}$ and $D^{I}_{aa}L=\{\varepsilon,bbba\}$.

In the special case $I=\emptyset$, we have $R^{\emptyset}_{\varepsilon}L=L$, $R^{\emptyset}_{u}L=\{\varepsilon\mid L{\downarrow}\}$ for any $u\neq\varepsilon$, and $D^{\emptyset}_{u}L=D_{u}L$. In the general case, the reorderable part and reordering derivative enjoy the following properties.

Lemma 4.3**.**

For every $L$, $L^{\prime}$, for any $u\in\Sigma^{*}$, if $L\subseteq L^{\prime}$, then $R^{I}_{u}L\subseteq R^{I}_{u}L^{\prime}$ and $D^{I}_{u}L\subseteq D^{I}_{u}L^{\prime}$.

Lemma 4.4**.**

For every $L$,

$R^{I}_{\varepsilon}L=L$*; for every $u,v\in\Sigma^{*}$, $R^{I}_{v}(R^{I}_{u}L)=R^{I}_{uv}L$; * 2. 2.

for every $u,u^{\prime}\in\Sigma^{*}$, $R^{I}_{\Sigma(u)}L=R^{I}_{\Sigma(u^{\prime})}L$.

We extend $R^{I}$ to subsets of $\Sigma$: by $R^{I}_{X}L$, we mean $R^{I}_{u}L$ where $u$ is any enumeration of $X$.

Lemma 4.5**.**

For every $L$,

$D^{I}_{\varepsilon}L=L$*; for any $u,v\in\Sigma^{*}$, $D^{I}_{v}(D^{I}_{u}L)=D^{I}_{uv}L$; * 2. 2.

for any $u,u^{\prime}\in\Sigma^{*}$ such that $u\sim^{I}u^{\prime}$, we have $D^{I}_{u}L=D^{I}_{u^{\prime}}L$.

Proposition 4.6**.**

For every $L$,

*for any $u\in\Sigma^{*}$, $D_{u}([L]^{I})=[D^{I}_{u}L]^{I}$; *

if $L$ is closed (i.e., $[L]^{I}=L$), then, for any $u\in\Sigma^{*}$, $D^{I}_{u}L$ is closed and $D_{u}L=D^{I}_{u}L$; 2. 2.

for any $u,v\in\Sigma^{*}$, $uv\in[L]^{I}$ iff $v\in[D^{I}_{u}L]^{I}$; 3. 3.

for any $u\in\Sigma^{*}$, $u\in[L]^{I}$ iff $(D^{I}_{u}L){\downarrow}$; 4. 4.

$[L]^{I}=\{\varepsilon\mid L{\downarrow}\}\cup\bigcup_{a\in\Sigma}\{a\}\cdot[D^{I}_{a}L]^{I}$.

Example 4.7**.**

Let $\Sigma=_{\mathrm{df}}\{a,b\}$ and $aIb$. Take $L$ to be the regular language $\llbracket(ab)^{*}\rrbracket$. We have already noted that the language $[L]^{I}=\{u\mid|u|_{a}=|u|_{b}\}$ is not regular. For any $n\in\mathbb{N}$, $D^{I}_{b^{n}}L=\{a^{n}\}\cdot L=\llbracket a^{n}(ab)^{*}\rrbracket$ whereas $D_{b^{n}}([L]^{I})=\{a^{n}\}\cdot^{I}[L]^{I}=\{u\mid|u|_{a}=|u|_{b}+n\}$. We can see that $[L]^{I}$ has infinitely many derivatives, none of which are regular, and $L$ has infinitely many reordering derivatives, all regular.

4.2 Brzozowski Reordering Derivative

The reorderable parts and reordering derivatives of regular languages turn out to be regular. We now show that they can be computed syntactically, generalizing the classical syntactic nullability and Brzozowski derivative operations [7].

Definition 4.8**.**

The $I$-reorderable part and the Brzozowski $I$-reordering derivative of a regexp are given by functions $R^{I},D^{I}:\mathsf{RE}\times\Sigma\to\mathsf{RE}$ and $R^{I},D^{I}:\mathsf{RE}\times\Sigma^{*}\to\mathsf{RE}$ defined recursively by

[TABLE]

The regexp $R_{u}E$ is nothing but $E$ with all occurrences of letters dependent with $u$ replaced with [math]. The definition of $D$ is more interesting. Compared to the classical Brzozowski derivative, the nullability condition $E{\downarrow}$ in the $EF$ case has been replaced with concatenation with the reorderable part $R^{I}_{a}E$, and the $E^{*}$ case has also been adjusted.

The functions $R$ and $D$ on regexps compute their semantic counterparts on the corresponding regular languages.

Proposition 4.9**.**

For any $E$,

for any $a\in\Sigma$, $R^{I}_{a}\llbracket E\rrbracket=\llbracket R^{I}_{a}E\rrbracket$ and $D^{I}_{a}\llbracket E\rrbracket=\llbracket D^{I}_{a}E\rrbracket$; 2. 2.

for any $u\in\Sigma^{*}$, $R^{I}_{u}\llbracket E\rrbracket=\llbracket R^{I}_{u}E\rrbracket$ and $D^{I}_{u}\llbracket E\rrbracket=\llbracket D^{I}_{u}E\rrbracket$.

Proposition 4.10**.**

For any $E$,

for any $a\in\Sigma$, $v\in\Sigma^{*}$, $av\in\llbracket E\rrbracket^{I}\iff v\in\llbracket D^{I}_{a}E\rrbracket^{I}$; 2. 2.

for any $u,v\in\Sigma^{*}$, $uv\in\llbracket E\rrbracket^{I}\iff v\in\llbracket D^{I}_{u}E\rrbracket^{I}$; 3. 3.

for any $u\in\Sigma^{*}$, $u\in\llbracket E\rrbracket^{I}\iff(D^{I}_{u}E){\downarrow}$.

Example 4.11**.**

Let $\Sigma=_{\mathrm{df}}\{a,b\}$, $aIb$ and $E=_{\mathrm{df}}aa+ab+b$.

[TABLE]

As with the classical Brzozowski derivative, we can use the reordering Brzozowski derivative to construct deterministic automata. For a regexp $E$, take $Q^{E}=_{\mathrm{df}}\{D^{I}_{u}E\mid u\in\Sigma^{*}\}$, $q^{E}_{0}=_{\mathrm{df}}E$, $F^{E}=_{\mathrm{df}}\{E^{\prime}\in Q^{E}\mid E^{\prime}{\downarrow}\}$, $\delta^{E}_{a}E^{\prime}=_{\mathrm{df}}D^{I}_{a}E^{\prime}$ for $E^{\prime}\in Q^{E}$. By Prop. 4.10, this automaton accepts the closure $\llbracket E\rrbracket^{I}$. But even quotiented by the full Kleene algebra theory, the quotient of $Q^{E}$ is not necessarily finite, i.e., we may be able to construct infinitely many different languages by taking reordering derivatives. For the regexp from Example 4.7, we have $D^{I}_{b^{n}}((ab)^{*})\doteq a^{n}(ab)^{*}$, so it has infinitely many Brzozowski reordering derivatives even up to the Kleene algebra theory. This is only to be expected, as the closure $\llbracket(ab)^{*}\rrbracket^{I}$ is not regular and cannot possibly have an accepting finite automaton.

4.3 Antimirov Reordering Derivative

Like the classical Brzozowski derivative that was optimized by Antimirov [3], the Brzozowski reordering derivative construction can be optimized by switching from functions on regexps to multivalued functions or relations.

Definition 4.12**.**

The Antimirov $I$-reordering parts-of-derivatives of a regexp along a letter and a word are relations ${\to^{I}}\subseteq\mathsf{RE}\times\Sigma\times\mathsf{RE}$ and ${\to^{I*}}\subseteq\mathsf{RE}\times\Sigma^{*}\times\mathsf{RE}$ defined inductively by

[TABLE]

Here $R^{I}$ is defined as before. Similarly to the Brzozowski reordering derivative from the previous subsection, the condition $E{\downarrow}$ in the second $EF$ rule has has been replaced with concatenation with $R^{I}_{a}E$, and the $E^{*}$ rule has been adjusted.

Collectively, the Antimirov reordering parts-of-derivatives of a regexp $E$ compute the semantic reordering derivative of the language $\llbracket E\rrbracket$.

Proposition 4.13**.**

For any $E$,

for any $a\in\Sigma$, $D^{I}_{a}\llbracket E\rrbracket=\bigcup\{\llbracket E^{\prime}\rrbracket\mid E\to^{I}(a,E^{\prime})\}$; 2. 2.

*for any $u\in\Sigma^{*}$, $D^{I}_{u}\llbracket E\rrbracket=\bigcup\{\llbracket E^{\prime}\rrbracket\mid E\to^{I*}(u,E^{\prime})\}$. *

Proposition 4.14**.**

For any $E$,

for any $a\in\Sigma$, $v\in\Sigma^{*}$, $av\in\llbracket E\rrbracket^{I}\iff\exists E^{\prime}.\,E\to^{I}(a,E^{\prime})\wedge v\in\llbracket E^{\prime}\rrbracket^{I}$; 2. 2.

for any $u,v\in\Sigma^{*}$, $uv\in\llbracket E\rrbracket^{I}\iff\exists E^{\prime}.\,E\to^{I*}(u,E^{\prime})\wedge v\in\llbracket E^{\prime}\rrbracket^{I}$; 3. 3.

for any $u\in\Sigma^{*}$, $u\in\llbracket E\rrbracket^{I}\iff\exists E^{\prime}.\,E\to^{I*}(u,E^{\prime})\wedge E^{\prime}{\downarrow}$.

Example 4.15**.**

Let us revisit Example 4.11. The Antimirov reordering parts-of-derivatives of $E$ along $b$ are $a1$ and $1$:

[TABLE]

The Antimirov reordering parts-of-derivatives of $E^{*}$ along $b$ are therefore $E_{b}^{*}(a1)E^{*}$ and $E_{b}^{*}1E^{*}$ where $E_{b}=_{\mathrm{df}}R^{I}_{b}E=aa+a0+0$. Recall that, for the Brzozowski reordering derivative, we computed $D^{I}_{b}E=(0a+a0)+(0b+a1)+1$ and $D^{I}_{b}E^{*}=E_{b}^{*}((0a+a0)+(0b+a1)+1)E^{*}$.

Like the classical Antimirov construction, the Antimirov reordering parts-of-derivatives of a regexp $E$ give a nondeterministic automaton by $Q^{E}=_{\mathrm{df}}\{E^{\prime}\mid\exists u\in\Sigma^{*}.\,E\to^{I*}(u,E^{\prime})\}$, $I^{E}=_{\mathrm{df}}\{E\}$, $F^{E}=_{\mathrm{df}}\{E^{\prime}\in Q^{E}\mid E^{\prime}{\downarrow}\}$, $E^{\prime}\to^{E}(a,E^{\prime\prime})=_{\mathrm{df}}E^{\prime}\to^{I}(a,E^{\prime\prime})$ for $E^{\prime},E^{\prime\prime}\in Q^{E}$. This automaton accepts $\llbracket E\rrbracket^{I}$ by Prop. 4.14, but is generally infinite, also if quotiented by the full Kleene algebra theory. Revisiting Example 4.7 again, $(ab)^{*}$ must have infinitely many Antimirov reordering parts-of-derivatives modulo the Kleene algebra theory since $\llbracket(ab)^{*}\rrbracket^{I}$ is not regular and cannot have a finite accepting nondeterministic automaton. Specifically, it has $(a0)^{*}((a1)\ldots((a0)^{*}((a1)(ab)^{*}))\ldots)\doteq a^{n}(ab)^{*}$ as its single reordering part-of-derivative along $b^{n}$.

However, if quotienting the Antimirov automaton for $E$ by some sound theory (a theory weaker than the Kleene algebra theory) makes it finite, then the Brzozowski automaton can also be quotiented to become finite.

Proposition 4.16**.**

For any $E$,

for any $a\in\Sigma$, $D^{I}_{a}E\doteq\sum\{E^{\prime}\mid E\to^{I}(a,E^{\prime})\}$; 2. 2.

for any $u\in\Sigma^{*}$, $D^{I}_{u}E\doteq\sum\{E^{\prime}\mid E\to^{I*}(u,E^{\prime})\}$

(using the semilattice equations for $0,+$, that [math] is zero, and distributivity of $\cdot$ over $+$).

Corollary 4.17**.**

If some quotient of the Antimirov automaton for $E$ (accepting $\llbracket E\rrbracket^{I}$) is finite, then also some quotient of the Brzozowski automaton is finite.

4.4 Star-Connected Expressions

Star-connected expressions are important as they characterize regular closed languages. A corollary of that is a further characterization of such languages in terms of a “concurrent” semantics of regexps that interprets Kleene star nonstandardly as “concurrent star”.

Definition 4.18**.**

A word $w\in\Sigma^{*}$ is connected if its dependence graph $\langle w\rangle_{D}$ is connected. A language $L\subseteq\Sigma^{*}$ is connected if every word $w\in L$ is connected.

Definition 4.19**.**

Star-connected expressions* are a subset of the set of all regexps defined inductively by: [math], $1$ and $a\in\Sigma$ are star-connected. If $E$ and $F$ are star-connected, then so are $E+F$ and $EF$. If $E$ is star-connected and $\llbracket E\rrbracket$ is connected, then $E^{*}$ is star-connected.* 2. 2.

A language $L$ is said to be star-connected if $L=\llbracket E\rrbracket$ for some star-connected regexp.

Ochmański [18] proved that a closed language is regular iff it is the closure of a star-connected language. This means that, for any regexp $E$, the language $\llbracket E\rrbracket^{I}$ is regular iff there exists a star-connected expression $E^{\prime}$ such that $\llbracket E\rrbracket^{I}=\llbracket E^{\prime}\rrbracket^{I}$. It is important to realize that generally $E\neq E^{\prime}$ and also $\llbracket E\rrbracket\neq\llbracket E^{\prime}\rrbracket$. Ochmański’s proof was as follows.

For a linear order $\leq$ on $\Sigma$, a word $z$ is a lexicographic normal form if $\forall w\in[z]^{I}.\,z\leq_{\mathrm{lex}}w$ where $\leq_{\mathrm{lex}}$ is the lexicographic order on $\Sigma^{*}$ induced by $\leq$. We write $\mathrm{Lex}^{I}$ for the set of all lexicographic normal forms.

Lemma 4.20**.**

(cf. [19, Props. 6.3.4, 6.3.10])

$\mathrm{Lex}^{I}$* is regular.* 2. 2.

For any regular expression $E$, if $\llbracket E\rrbracket\subseteq\mathrm{Lex}^{I}$, then $E$ is star-connected.

Theorem 4.21** (Ochmański [18]).**

(cf. [19, Thm. 6.3.13]) For any closed language $L$, the following are equivalent:

$L$* is regular;* 2. 2.

$L\cap\mathrm{Lex}^{I}$* is regular;* 3. 3.

$L$* is star-connected.*

Proof 4.22**.**

(1) $\Rightarrow$ (2) is a consequence of Lemma 4.20(1) as the intersection of regular languages is regular. (2) $\Longrightarrow$ (3) follows from Lemma 4.20(2). For (3) $\Longrightarrow$ (1), Ochmański employed Hachiguchi’s notion of rank of a language and Hachiguchi’s lemma, which we will study in Def. 5.1 and Prop. 5.2 below, and proved that, if $L$ is closed and connected, then $L^{*}$ has rank.

The nonstandard concurrent-star trace-language semantics of regular expressions $\llbracket\_\rrbracket^{\mathrm{con}}:\mathsf{RE}(\Sigma)\to\mathcal{P}\Sigma^{*}$ is like $\llbracket\_\rrbracket$ except that the star constructor is interpreted nonstandardly as the concurrent star operation. Informally, the concurrent star of a language iterates not the given language but the language of connected components of its words.

The concurrent star of a connected language coincides with its Kleene star. The idea of this nonstandard semantics is to make non-star-connected regular expressions harmless, so as to obtain the following replacement for Kleene’s theorem.

Theorem 4.23** (Ochmański [18]).**

(cf. [19, Thm. 6.3.16]) A closed langugage $L$ is regular iff $L=[\llbracket E\rrbracket^{\mathrm{con}}]^{I}$ for some regexp $E$.

4.5 Automaton Finiteness for Star-Connected Expressions

We now show that the set of Antimirov reordering parts-of-derivatives of a star-connected expression is finite modulo suitable equations.

Lemma 4.24**.**

If a language $L$ is connected, then for any $u\in\Sigma^{+}$, $R^{I}_{u}(D^{I}_{u}L)\subseteq\mathbf{1}$.

Proof 4.25**.**

Because $L$ is connected, if $w\in D^{I}_{u}L$, then $aDb$ for some $a\in u$ and $b\in w$. For such $w$ to also be in $R^{I}_{u}(D^{I}_{u}L)$, we also need that $wIu$. This is only possible if $w=\varepsilon$.

Lemma 4.26**.**

For any $E$, if $\llbracket E\rrbracket\subseteq\mathbf{1}$, then either $E\doteq 0$ or $E\doteq 1$ (using the equations involving [math] and $1$ only (e.g., $0+1\doteq 1$ and $0^{*}\doteq 1$ etc.) and that [math] is zero).

Lemma 4.27**.**

For any $E$, $E^{\prime}$ and $u\in\Sigma^{+}$, if $\llbracket E\rrbracket$ is connected and $E\to^{I*}(u,E^{\prime})$, then $R^{I}_{u}E^{\prime}\doteq 0$ or $R^{I}_{u}E^{\prime}\doteq 1$ (using the equations involving [math] and $1$ only and that [math] is zero).

Proof 4.28**.**

From $E\to^{I*}(u,E^{\prime})$ by Proposition 4.13, $\llbracket E^{\prime}\rrbracket\subseteq D^{I}_{u}\llbracket E\rrbracket$. Hence by Lemma 4.24, we get $\llbracket R^{I}_{u}E^{\prime}\rrbracket=R^{I}_{u}\llbracket E^{\prime}\rrbracket\subseteq R^{I}_{u}(D^{I}_{u}\llbracket E\rrbracket)\subseteq\mathbf{1}$. By Lemma 4.26, $R^{I}_{u}E^{\prime}\doteq 0$ or $R^{I}_{u}E^{\prime}\doteq 1$.

Lemma 4.29**.**

For any $E$, $E^{\prime}$ and $u\in\Sigma^{*}$, if $E^{*}\to^{I*}(u,E^{\prime})$, then there exist $n\in\mathbb{N}$, $E_{1},\ldots,E_{n}$, $\emptyset\subset X_{0},\ldots,X_{n-1}\subseteq\Sigma$ and $u_{1},\ldots,u_{n}\in\Sigma^{+}$ such that

[TABLE]

where $X_{i-1}\supseteq X_{i}\cup\Sigma(u_{i})$ and $E\to^{I*}(u_{i},E_{i})$ for all $i$ (using only associativity of $\cdot$).

Lemma 4.30**.**

For any $E$, $E^{\prime}$ and $u\in\Sigma^{*}$, if $\llbracket E\rrbracket$ is connected, $E^{*}\to^{I*}(u,E^{\prime})$ and, for the development of $E^{\prime}$ from the previous lemma, we have $X_{i-1}=X_{i}$ for some $i$, then $R^{I}_{X_{i}}E_{i}\doteq 0$ or $R^{I}_{X_{i}}E_{i}\doteq 1$ (using the equations involving [math] and $1$ only, that [math] is zero).

Proof 4.31**.**

We have $\Sigma(u_{i})\subseteq X_{i-1}=X_{i}$. From $E\to^{I*}(u_{i},E_{i})$, by Lemma 4.27 either $R^{I}_{u_{i}}E_{i}\doteq 0$ or $R^{I}_{u_{i}}E_{i}\doteq 1$. Therefore also $R^{I}_{X_{i}}E_{i}\doteq 0$ or $R^{I}_{X_{i}}E_{i}\doteq 1$.

Lemma 4.32**.**

For any $E$, $E^{\prime}$ and $u\in\Sigma^{*}$, if $\llbracket E\rrbracket$ is connected and $E^{*}\to^{I*}(u,E^{\prime})$, then there exist $n\leq|\Sigma|$, $E_{1},\ldots,E_{n}$ and $\emptyset\subset X_{0},\ldots,X_{n-1}\subseteq\Sigma$ such that

[TABLE]

and $X_{i-1}\supset X_{i}$ for all $i$ (using, in addition to the equations mentioned in the lemmata above, unitality of $1$ and the equation $F^{*}\cdot F^{*}\doteq F^{*}$).

Proof 4.33**.**

From Lemmata 4.29, 4.30 noting that at most $|\Sigma|-1$ of the inclusions $X_{i-1}\supseteq X_{i}$ can be proper.

Definition 4.34**.**

We define functions $(\_)^{\to+},(\_)^{\to*}:\mathsf{RE}\to\mathcal{P}\mathsf{RE}$ by

[TABLE]

Proposition 4.35**.**

For any $E$, $E^{\prime}$ and $u\in\Sigma^{*}$, if $E\to^{I*}(u,E^{\prime})$, then there exists $E^{\prime\prime}$ such that $E^{\prime}\doteq E^{\prime\prime}$ and $E^{\prime\prime}\in E^{\to*}$ (using only the equations mentioned in the above lemmata).

Proposition 4.36**.**

If $E$ is star-connected, then a suitable sound quotient of the state set $\{E^{\prime}\mid\exists u\in\Sigma^{*}.\ E\to^{I*}(u,E^{\prime})\}$ of the Antimirov automaton for $E$ (accepting $\llbracket E\rrbracket^{I}$) is finite.

Proof 4.37**.**

By Lemma 4.32, for a star-connected expression $E$, we only need to consider $n\leq|\Sigma|$ in the definition of $(E^{*})^{\to+}$ for Proposition 4.35 to hold. This restriction makes the set $E^{\to*}$ finite.

5 Uniform Scattering Rank of a Language

We proceed to defining the notion of uniform scattering rank of a language and show that star-connected expressions define languages with uniform scattering rank.

5.1 Scattering Rank vs. Uniform Scattering Rank

The notion of scattering rank of a language (a.k.a. distribution rank, $k$-block testability) was introduced by Hashiguchi [9].

Definition 5.1**.**

*A language $L$ has ($I$-scattering) rank at most $N$ if

$\forall u,v.\,uv\in[L]^{I}\Longrightarrow\exists z\in L.\,u\mathrel{{\sim}{\lhd}}_{N}z\mathrel{{\rhd}{\sim}}v$.*

We say that $L$ has rank if it has rank at most $N$ for some $N\in\mathbb{N}$. If it does, then, for the least such $N$, we say that $L$ has rank $N$.

The only languages with rank 0 are $\emptyset$ and $\mathbf{1}$. If a nontrivial language $L$ is closed, it has rank 1: for any $uv\in[L]^{I}$, we have also have $uv\in L$ and $u\lhd uv\rhd\varepsilon,v$.

Having rank is a sufficient condition for regularity of the trace closure of a regular language. But it is not a necessary condition.

Proposition 5.2** (Hashiguchi [9]).**

(cf. [19, Prop. 6.3.2]) If a regular language $L$ has rank, then $[L]^{I}$ is regular.

Proposition 5.3**.**

There exist regular languages $L$ such that $[L]^{I}$ is regular but $L$ is without a rank.

Proof 5.4**.**

Consider $\Sigma=_{\mathrm{df}}\{a,b\}$, $aIb$. The regular language $L=_{\mathrm{df}}\llbracket(ab)^{*}(a^{*}+b^{*})\rrbracket$ is without a rank, since, for any $n$, we have $(ab)^{n}\in L$ and $a^{n}b^{n}\in[L]^{I}$ while the smallest $N$ such that $a^{n}\mathrel{{\sim}{\lhd}}_{N}(ab)^{n}\mathrel{{\rhd}{\sim}}b^{n}$ is $n$. Nonetheless, $[L]^{I}=\Sigma^{*}=\llbracket(a+b)^{*}\rrbracket$ is regular.

We wanted to show that a truncation of the refined Antimirov automaton (which we define in Section 6) is finite for regexps whose language has rank. But it turns out, as we shall see, that rank does not quite work for this. For this reason, we introduce a stronger notion that we call uniform scattering rank.

Definition 5.5**.**

*A language $L$ has uniform ($I$-scattering) rank at most $N$ if

$\forall w\in[L]^{I}.\ \exists z\in L.\ \forall u,v.\,w=uv\Longrightarrow u\mathrel{{\sim}{\lhd}}_{N}z\mathrel{{\rhd}{\sim}}v$.*

The difference between the two definitions is that, in the uniform case, the choice of $z$ depends only on $w$ whereas, in the non-uniform case, it depends on the particular split of $w$ as $w=uv$, i.e., for every such split of $w$ we may choose a different $z$.

Lemma 5.6**.**

If $L$ has uniform rank at most $N$, then $L$ has rank at most $N$.

The converse of the above lemma does not hold—there are languages with uniform rank greater than rank. Furthermore, there are languages that have rank but no uniform rank.

Proposition 5.7**.**

Let $\Sigma=_{\mathrm{df}}\{a,b,c\}$, $aIb$ and $E=_{\mathrm{df}}a^{*}b^{*}c(ab)^{*}(a^{*}+b^{*})+(ab)^{*}(a^{*}+b^{*})ca^{*}b^{*}$.

The language $\llbracket E\rrbracket$ has rank 2. 2. 2.

The language $\llbracket E\rrbracket$ has no uniform rank.

Proof 5.8**.**

Note that $c$ behaves like a separator—although $a$ and $b$ are independent, neither $a$ nor $b$ commutes with $c$. It can be seen that words in $\llbracket E\rrbracket^{I}$ are of the form $w_{l}cw_{r}$ where $w_{l}$ and $w_{r}$ consist of some number of $a$’s and $b$’s, i.e., $\llbracket E\rrbracket^{I}=\llbracket(a+b)^{*}c(a+b)^{*}\rrbracket$.

Let $uv\in\llbracket E\rrbracket^{I}$. We have to find $u_{1},u_{2}$ and $v_{0},v_{1},v_{2}$ so that $u_{1}u_{2}\sim^{I}u$, $v_{0}v_{1}v_{2}\sim^{I}v$, $v_{0}Iu_{1}$, $v_{0}v_{1}Iu_{2}$ and $v_{0}u_{1}v_{1}u_{2}v_{2}\in\llbracket E\rrbracket$. There are two cases to consider: either $c$ is in the suffix $v$ or it is in the prefix $u$.

• •

Case $c\in v$: We have that $u$ consists of only $a$’s and $b$’s. Let $x,y\in\Sigma^{*}$ be such that $v=xcy$. Set $u_{1}=_{\mathrm{df}}\pi_{a}(u)$, $u_{2}=_{\mathrm{df}}\pi_{b}(u)$, $v_{0}=_{\mathrm{df}}\varepsilon$ and $v_{1}=_{\mathrm{df}}\pi_{a}(x)$. Let $k=_{\mathrm{df}}|y|_{a}$, $l=_{\mathrm{df}}|y|_{b}$ and $m=_{\mathrm{df}}\min(k,l)$. Set $v_{2}=_{\mathrm{df}}\pi_{b}(x)c(ab)^{m}a^{k-m}b^{l-m}$. We have that $u_{1}u_{2}=\pi_{a}(u)\pi_{b}(u)\sim^{I}u$. Since $\pi_{a}(x)\pi_{b}(x)\sim^{I}x$ and $(ab)^{m}a^{k-m}b^{l-m}\sim^{I}y$, we also have $v_{0}v_{1}v_{2}=\varepsilon\pi_{a}(x)\pi_{b}(x)c(ab)^{m}a^{k-m}b^{l-m}\sim^{I}xcy=v$. Also, $\varepsilon I\pi_{a}(u)$ and $\varepsilon\pi_{a}(x)I\pi_{b}(u)$. Finally, $v_{0}u_{1}v_{1}u_{2}v_{2}=\varepsilon\pi_{a}(u)\pi_{a}(x)\pi_{b}(u)\pi_{b}(x)c(ab)^{m}a^{k-m}b^{l-m}\in\llbracket a^{*}b^{*}c(ab)^{*}(a^{*}+b^{*})\rrbracket$.

• •

Case $c\in u$: Similar to the previous case. Let $x$ and $y$ be such that $u=xcy$. Let $k=_{\mathrm{df}}|x|_{a}$, $l=_{\mathrm{df}}|x|_{b}$ and $m=_{\mathrm{df}}\min(k,l)$. Set $u_{1}=_{\mathrm{df}}(ab)^{m}a^{k-m}b^{l-m}c\pi_{a}(y)$, $u_{2}=_{\mathrm{df}}\pi_{b}(y)$, $v_{0}=_{\mathrm{df}}\varepsilon$, $v_{1}=_{\mathrm{df}}\pi_{a}(v)$ and $v_{2}=_{\mathrm{df}}\pi_{b}(v)$. In this case we have $v_{0}u_{1}v_{1}u_{2}v_{2}\in\llbracket(ab)^{*}(a^{*}+b^{*})ca^{*}b^{*}\rrbracket$. 2. 2.

Assume that $\llbracket E\rrbracket$ has uniform rank at most $N$. Take $w=_{\mathrm{df}}a^{N+1}b^{N+1}ca^{N+1}b^{N+1}\in\llbracket E\rrbracket^{I}$. By our assumption, there is $z\in\llbracket E\rrbracket$ such that, for all $u$ and $v$, if $w=uv$, then $u\mathrel{{\sim}{\lhd}}_{N}z\mathrel{{\rhd}{\sim}}v$.

Take $u=_{\mathrm{df}}a^{N+1}$ and $v=_{\mathrm{df}}b^{N+1}ca^{N+1}b^{N+1}$. Thus, for some $n\leq N$, $z=v_{0}u_{1}v_{1}\ldots u_{n}v_{n}$ and $u=a^{N+1}\sim^{I}u_{1}\ldots u_{n}$. Since we have $N+1$ letters $a$ to divide into $n\leq N$ words, at least one $u_{i}$ must consist of more than one $a$ and thus $z$ must contain two consecutive $a$’s that are before $c$.

Take $u^{\prime}=_{\mathrm{df}}a^{N+1}b^{N+1}ca^{N+1}$ and $v^{\prime}=_{\mathrm{df}}b^{N+1}$. Again, for some $n\leq N$, $z=v^{\prime}_{0}u^{\prime}_{1}v^{\prime}_{1}\ldots u^{\prime}_{n}v^{\prime}_{n}$ and $v^{\prime}=b^{N+1}\sim^{I}v^{\prime}_{0}\ldots v^{\prime}_{n}$. Note that $c$ must be in one of the $u^{\prime}_{i}$’s and thus for all $j<i$ it must be that $v^{\prime}_{j}=\varepsilon$. Hence $v^{\prime}_{0}=\varepsilon$ and we have $N+1$ letters $b$ to divide into $n\leq N$ words and thus at least one $v^{\prime}_{i}$ consists of more than one $b$. This means that $z$ must contain two consecutive $b$’s that are after $c$.

The only words in $\llbracket E\rrbracket$ equivalent to $w$ are $a^{N+1}b^{N+1}c(ab)^{N+1}$ and $(ab)^{N+1}ca^{N+1}b^{N+1}$. Neither of these has at least two consecutive $a$’s before $c$ as well as at least two consecutive $b$’s after $c$, so neither qualifies as $z$. Contradiction.

5.2 Star-Connected Languages Have Uniform Rank

Klunder et al. [12] established that star-connectedness is a sufficient condition for a regular language to have rank, although not a necessary one.

Proposition 5.9** **(Klunder et

al. [12]).

Any star-connected language has rank.

Proof 5.10**.**

The language $\{a\}$ has rank $1$. The languages $\emptyset$ and $\mathbf{1}$ have rank [math]. If two languages $L_{1}$ and $L_{2}$ have ranks at most $N_{1}$ resp. $N_{2}$, then $L_{1}\cup L_{2}$ has rank at most $\max(N_{1},N_{2})$ and $L_{1}\cdot L_{2}$ has rank at most $N_{1}+N_{2}$. If a general language $L$ has rank at most $N$, then $L^{*}$ need not have rank. For example, for $\Sigma=_{\mathrm{df}}\{a,b\}$, $aIb$, the language $\{ab\}$ has rank 1, but $\{ab\}^{*}$ is without rank. But if $L$ is also connected, then $L^{*}$ turns out to have rank at most $(N+1)\cdot|\Sigma|$. The claim follows by induction on the given star-connected expression.

Proposition 5.11**.**

There exist regular languages with rank (and also with uniform rank) that are not star-connected.

Proof 5.12**.**

Consider $\Sigma=_{\mathrm{df}}\{a,b\}$, $aIb$. The language $L=_{\mathrm{df}}\llbracket(aa+ab+ba+bb)^{*}\rrbracket$ has rank 1, in fact even uniform rank 1, because it is closed. The regular expression $(aa+ab+ba+bb)^{*}$ is clearly not star-connected, since the language $\llbracket aa+ab+ba+bb\rrbracket$ contains disconnected words $ab$ and $ba$. But a more involved pumping argument also shows that $L$ is not star-connected, i.e., that there is no star-connected expression $E$ such that $L=\llbracket E\rrbracket$.

We will now show that star-connected languages also have uniform rank, by refining Klunder et al.’s proof of Proposition 5.9, especially the case of the Kleene star.

Let us analyze the case $L^{*}$ where $L$ is a connected language. When $w\in[L^{*}]^{I}$, then there exists $z\in L$ such that $w\sim^{I}z$. This further means that there exist $n\in N$ and $z_{1},\ldots,z_{n}\in L$ such that $z=z_{1}\ldots z_{n}$ where we can require that all $z_{i}$ are nonempty. Since $L$ is connected, each $z_{i}$ is also connected. If $w=uv$, then there exist $u_{1},\ldots,u_{n}$ and $v_{1},\ldots,v_{n}$ such that $u\sim^{I}u_{1}\ldots u_{n}$, $v\sim^{I}v_{1}\ldots v_{n}$ and, for every $i$, $z_{i}\sim^{I}u_{i}v_{i}$ and, for every $j<i$, $v_{j}Iu_{i}$. In other words, $u_{i}$ is the part of $z_{i}$ that belongs to $u$ and $v_{i}$ is the part that belongs to $v$. In particular, if $u_{i}=\varepsilon$ (or $z_{i}\sim^{I}v_{i}$), then all letters of $z_{i}$ belong to the suffix $v$, and similarly if $v_{i}=\varepsilon$ (or $z_{i}\sim^{I}u_{i}$), then all letters of $z_{i}$ belong to the prefix $u$. An important observation for us is that not more than $|\Sigma|$ of the $z_{i}$ can be two-colored in the sense that both $u_{i}\neq\varepsilon$ and $v_{i}\neq\varepsilon$.

Lemma 5.13**.**

Let $w,u,v,z_{1},\ldots,z_{n}$ be words such that $w=uv$, $w\sim^{I}z_{1}\ldots z_{n}$ and each $z_{i}$ is nonempty and connected. Let $u_{1},\ldots,u_{n},v_{1},\ldots,v_{n}$ be words such that $u\sim^{I}u_{1}\ldots u_{n}$, $v\sim^{I}v_{1}\ldots v_{n}$, for all $i$, $z_{i}\sim^{I}u_{i}v_{i}$, and, for all $j<i$, $v_{j}Iu_{i}$. For at most $|\Sigma|$ of the words $z_{i}$, it can be that both $u_{i}\neq\varepsilon$ and $v_{i}\neq\varepsilon$.

Proof 5.14**.**

If, for some $i$, we have that $u_{i}\neq\varepsilon$ and $v_{i}\neq\varepsilon$, then, since $z_{i}\sim^{I}u_{i}v_{i}$ is connected, there must exist letters $a$ and $b$ such that $a\in u_{i}$, $b\in v_{i}$ and $aDb$. Since $v_{i}Iu_{i+1}\ldots u_{n}$, we have $a\not\in u_{i+1}\ldots u_{n}$. This means that, if there are $k$ words $z_{i}$ such that $u_{i}\neq\varepsilon$ and $v_{i}\neq\varepsilon$, then these words together must contain at least $k$ distinct letters.

Should it happen for some $i$ that $z_{i}$ and $z_{i+1}$ are completely from the prefix $u$ (in the sense that $v_{i}=v_{i+1}=\varepsilon$, i.e., $z_{i}\sim^{I}u_{i}$ and $z_{i+1}\sim^{I}u_{i+1}$), then $z_{i}$ and $z_{i+1}$ belong to the same block of $u$ in the scattering $u\mathrel{{\sim}{\lhd}}z\mathrel{{\rhd}{\sim}}v$, which can potentially help keeping the uniform rank of $L^{*}$ low. The same holds for $z_{i}$ and $z_{i+1}$ that are completely from the suffix $v$: they belong to the same block of $v$. Having words $z_{i}$ that are completely from $u$ interspersed with other types of words $z_{i}$ (for example, having all odd-numbered $z_{i}$ completely from $u$ and all even-numbered are completely from $v$), in contrast, is not helpful. It could thus be useful to be able to choose $z$, $n$ and $z_{1}\ldots,z_{n}$ in such a way that as many as possible of the $z_{i}$ that are completely from $u$ are adjacent in $z$ for all splits of $w$ of as $w=uv$.

For example, take $\Sigma=_{\mathrm{df}}\{a,b\}$, $aIb$ and $L=_{\mathrm{df}}\Sigma=\llbracket a+b\rrbracket$. For $w=_{\mathrm{df}}a^{m}b^{m}\in[L^{*}]^{I}$, we could build $z=a^{m}b^{m}\in L^{*}$ from $n=_{\mathrm{df}}2m$, $z_{i}=_{\mathrm{df}}a$ for $1\leq i\leq m$ and $z_{i}=_{\mathrm{df}}b$ for $m+1\leq i\leq 2m$. Another option is to construct $z=(ab)^{m}$ from $n=_{\mathrm{df}}2m$, $z_{2i-1}=_{\mathrm{df}}a$ and $z_{2i}=_{\mathrm{df}}b$ for $1\leq i\leq m$. In the first case, the letters from $u$ stay together in $z$ for all prefixes $u$ of $w$ (as $w=z$). In the second case, they can be interleaved with the letters from $v$ (in the most extreme case $u=_{\mathrm{df}}a^{m}$ and $v=_{\mathrm{df}}b^{m}$, the words $u$ and $v$ get scattered into $m$ resp. $m+1$ blocks in $z$). Note that, when $z_{i}\sim^{I}v_{i}$ and $z_{i+1}\sim^{I}u_{i+1}$, then $z_{i}Iz_{i+1}$. The next lemma says that, for given $z$, $n$, $z_{1},\ldots,z_{n}$, the sequence of words $z_{1},\ldots,z_{n}$ can be permuted into $z^{\prime}_{1},\ldots,z^{\prime}_{n}$ with $z_{1}\ldots z_{n}\sim^{I}z^{\prime}_{1}\ldots z^{\prime}_{n}$ so that, if $z^{\prime}_{i}$ is completely from $u$, then $z^{\prime}_{i-1}$ is not completely from $v$. As $z_{i}\in L$ for all $i$, it is of course the case that $z^{\prime}=_{\mathrm{df}}z^{\prime}_{1}\ldots z^{\prime}_{n}\in L^{*}$, so $z^{\prime}$ is as good a witness of $w\in[L^{*}]^{I}$ as $z$.

Lemma 5.15**.**

Let $w,u,v,z_{1},\ldots,z_{n}$ be words such that $w=uv$, $w\sim^{I}z_{1}\ldots z_{n}$, and each $z_{i}$ is nonempty and connected. There exists a permutation $\sigma^{\prime}=z^{\prime}_{1},\ldots,z^{\prime}_{n}$ of $\sigma=_{\mathrm{df}}z_{1},\ldots,z_{n}$ with the following properties:

$z_{1}\ldots z_{n}\sim^{I}z^{\prime}_{1}\ldots z^{\prime}_{n}$; 2. 2.

for any $u^{\prime},v^{\prime}$ such that $u=u^{\prime}v^{\prime}$, and for any $u_{1},\ldots,u_{n},v_{1},\ldots v_{n}$ such that $u^{\prime}\sim^{I}u_{1}\ldots u_{n}$, $v^{\prime}v\sim^{I}v_{1}\ldots v_{n}$, for all $i$, $z^{\prime}_{i}\sim^{I}u_{i}v_{i}$, and, for all $j<i$, $v_{j}Iu_{i}$, we have: if $v_{i}=\varepsilon$, then $u_{i-1}\neq\varepsilon$ unless $i=1$.

Proof 5.16**.**

By induction on $u$.

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Case $\varepsilon$: The identical permutation $\sigma^{\prime}=_{\mathrm{df}}\sigma$ has property 1 trivially. It also enjoys property 2 since $\varepsilon=u^{\prime}v^{\prime}$ implies $u^{\prime}=v^{\prime}=\varepsilon$, and, for all $i$, we have $z_{i}\sim^{I}u_{i}v_{i}$, $z_{i}\neq\varepsilon$, $u_{i}=\varepsilon$ and hence $v_{i}\neq\varepsilon$.

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Case $ua$: By induction hypothesis, we have a permutation $\sigma^{\prime}=z^{\prime}_{1},\ldots,z^{\prime}_{n}$ of $\sigma$ which has property 1 and and also has property 2 for all prefixes $u^{\prime}$ of $w$ up to $u$. Now consider the case where $u^{\prime}=_{\mathrm{df}}ua$ and $v^{\prime}=_{\mathrm{df}}\varepsilon$. This particular $a$ is in one of the $z^{\prime}_{i}$, say $z^{\prime}_{m}$. The only difference with the case $u^{\prime}=_{\mathrm{df}}u$ and $v^{\prime}=_{\mathrm{df}}a$ is that this $a$ is now in the $u^{\prime}$ part of $z^{\prime}_{m}$ and no longer in the $v^{\prime}v$ part. Let us also note that every nonempty $u_{i}$ has remained nonempty.

If $v_{m}\neq\varepsilon$, then the empty $v_{i}$ are exactly the same as in the case $u^{\prime}=_{\mathrm{df}}u$. Hence $\sigma^{\prime}$ has property 2 also for the prefix $u^{\prime}=_{\mathrm{df}}ua$.

If $v_{m}=\varepsilon$, but $m=1$ or $u_{m-1}\neq\varepsilon$, then $\sigma^{\prime}$ has property 2 also for the prefix $u^{\prime}=_{\mathrm{df}}ua$.

*In the critical case $v_{m}=\varepsilon$, $m\neq 1$ and $u_{m-1}=\varepsilon$, we construct a new permutation $\sigma^{\prime\prime}=_{\mathrm{df}}z^{\prime\prime}_{1},\ldots,z^{\prime\prime}_{n}$ from $\sigma^{\prime}$ by moving the words $z^{\prime}_{m},\ldots,z^{\prime}_{l}$ (where $l$ is the largest such that $v_{m}\ldots v_{l}=\varepsilon$) in front of of $z^{\prime}_{k},\ldots,z^{\prime}_{m-1}$ (where $k$ is the smallest such that $u_{k}\ldots u_{m-1}=\varepsilon$). Moving all these words rather than just $z_{m}$ alone ensures that the new permutation $\sigma^{\prime\prime}$ has property 2 also for all prefixes $u^{\prime}$ up to $u^{\prime}=u$ and not just only for the prefix $u^{\prime}=ua$. The new permutation $\sigma^{\prime\prime}$ also has property 1: indeed, we have $z_{1}\ldots z_{n}\sim^{I}z^{\prime}_{1}\ldots z^{\prime}_{n}\sim^{I}z^{\prime\prime}_{1}\ldots z^{\prime\prime}_{n}$ as $z^{\prime}_{k}\ldots z^{\prime}_{m-1}\sim^{I}v_{k}\ldots v_{m-1}\leavevmode\nobreak\ I\leavevmode\nobreak\ u_{m}\ldots u_{l}\sim^{I}z^{\prime}_{m}\ldots z^{\prime}_{l}$. *

Proposition 5.17**.**

If $E$ is star-connected, then the language $\llbracket E\rrbracket$ has uniform rank.

Proof 5.18**.**

By induction on $E$. We only look at the case $E^{*}$.

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Case $E^{*}$: From the assumption we have that $E$ is star-connected and $\llbracket E\rrbracket$ is connected. By induction hypothesis $\llbracket E\rrbracket$ has uniform rank at most $N$ for some $N\in\mathbb{N}$. We show that $\llbracket E^{*}\rrbracket$ has uniform rank at most $(|\Sigma|+1)N$.

Let $w\in\llbracket E^{*}\rrbracket^{I}$. Then there exist unique $n$ and $w_{1},\ldots,w_{n}$ such that $w\in w_{1}\cdot^{I}\ldots\cdot^{I}w_{n}$, $w_{i}\in\llbracket E\rrbracket^{I}$, and we can also require that $w_{i}\neq\varepsilon$. By $\llbracket E\rrbracket$ having uniform rank at most $N$, for every $i$, there exists a nonempty word $z_{i}\in\llbracket E\rrbracket$ such that, for any split of $w_{i}$ as $u_{i}v_{i}$, we have $u_{i}\mathrel{{\sim}{\lhd}}_{N}z_{i}\mathrel{{\rhd}{\sim}}v_{i}$. By connectedness of $\llbracket E\rrbracket$, all $z_{i}$ are connected.

We take $z^{\circ}=_{\mathrm{df}}z^{\circ}_{1}\ldots z^{\circ}_{n}$ where $\sigma^{\circ}=_{\mathrm{df}}z^{\circ}_{1},\ldots,z^{\circ}_{n}$ is the permutation of $\sigma=z_{1},\ldots,z_{n}$ obtained by Lemma 5.15 for $u=_{\mathrm{df}}w$ and $v=_{\mathrm{df}}\varepsilon$, i.e., for the specific split of $w$ as $w\varepsilon$. By Lemma 5.15(1), we have $w\sim^{I}w_{1}\ldots w_{n}\sim^{I}z_{1}\ldots z_{n}\sim^{I}z^{\circ}_{1}\ldots z^{\circ}_{n}=z^{\circ}$. We let $w^{\circ}_{1},\ldots,w^{\circ}_{n}$ be the corresponding permutation of $w_{1},\ldots,w_{n}$, so we also have $w\sim^{I}w^{\circ}_{0}\ldots w^{\circ}_{n}$ and $w^{\circ}_{i}\sim^{I}z^{\circ}_{i}$ for all $i$.

We will now show that, for any split of $w$ as $w=uv$, we have $u\mathrel{{\sim}{\lhd}}_{(|\Sigma|+1)N}z^{\circ}\mathrel{{\rhd}{\sim}}v$.

Let $w=uv$ be any split of $w$. There exist unique $u_{1},\ldots,u_{n},v_{1},\ldots v_{n}$ such $u\sim^{I}u_{1}\ldots u_{n}$, $v\sim^{I}v_{1}\ldots v_{n}$, for all $i$, $w^{\circ}_{i}=u_{i}v_{i}$, and for all $j<i$, $v_{j}Iu_{i}$. They give us $u_{1}\ldots u_{n}\lhd z^{\circ}\rhd v_{1}\ldots v_{n}$ and thus $u\mathrel{{\sim}{\lhd}}z^{\circ}\mathrel{{\rhd}{\sim}}v$. By Lemma 5.15(2) for $u^{\prime}=_{\mathrm{df}}u$, $v^{\prime}=_{\mathrm{df}}\varepsilon$, we have that $v_{i}=\varepsilon$ implies $u_{i-1}\neq\varepsilon$ unless $i=1$. By Lemma 5.13, there can be at most $|\Sigma|$ words $z^{\circ}_{i}$ such that both $u_{i}\neq\varepsilon$ and $v_{i}\neq\varepsilon$.

Each of these two-colored $z^{\circ}_{i}$ contributes at most $N$ to the degree of $u\mathrel{{\sim}{\lhd}}z^{\circ}\mathrel{{\rhd}{\sim}}v$, so altogether they contribute at most $|\Sigma|N$.

Between any two-colored $z^{\circ}_{i}$ and also before the first and after the last one of them, there are some $z^{\circ}_{i}$ completely from $u$ followed by some $z^{\circ}_{i}$ completely from $v$. Each such sequence contributes at most 1 to the degree of $u\mathrel{{\sim}{\lhd}}z^{\circ}\mathrel{{\rhd}{\sim}}v$. If there are less than $|\Sigma|$ two-colored words, these sequences thus contribute altogether at most $|\Sigma|$ to the degree. If there are exactly $|\Sigma|$ two-colored words $z^{\circ}_{i}$, then the $z^{\circ}_{i}$ after the last of them are all completely from $v$, so their sequence belongs to the last $v$-block generated by the last two-colored $z^{\circ}_{i}$ and thus contributes 0. Again altogether these sequences contribute at most $|\Sigma|$.

*Altogether the degree of $u\lhd z^{\circ}\rhd v$ is thus at most $(|\Sigma|+1)N$. *

6 Antimirov Reordering Derivative and Uniform Rank

We have seen that the reordering language derivative $D^{I}_{u}L$ allows $u$ to be scattered in a word $z\in L$ as $u_{1},\ldots,u_{n}\lhd z\rhd v_{0},\ldots,v_{n}$ where $u\sim^{I}u_{1}\ldots u_{n}$. We will now consider a version of the Antimirov reordering derivative operation that delivers lists of regexps for the possible $v_{0},\ldots,v_{n}$ rather than just single regexps for their concatenations $v_{0}\ldots v_{n}$.

6.1 Refined Antimirov Reordering Derivative

The refined reordering parts-of-derivative of a regexp $E$ along a letter $a$ are pairs of regexps $E_{l},E_{r}$. For any word $w=av\in\llbracket E\rrbracket^{I}$, there must be an equivalent word $z=v_{l}av_{r}\in\llbracket E\rrbracket$. Instead of describing the words $v_{l}v_{r}$ obtainable by removing a minimal occurrence of $a$ in a word $z\in\llbracket E\rrbracket$, the refined parts-of-derivative describe the subwords $v_{l},v_{r}$ that were to the left and right of this $a$ in $z$: it must be the case that $v_{l}\in\llbracket E_{l}\rrbracket$ and $v_{r}\in\llbracket E_{r}\rrbracket$ for one of the pairs $E_{l},E_{r}$. For a longer word $u$, the refined reordering derivative operation gives lists of regexps $E_{0},\ldots,E_{n}$ fixing what the lists of subwords $v_{0},\ldots,v_{n}$ can be in words $z=v_{0}u_{1}v_{1}\ldots u_{n}v_{n}\in\llbracket E\rrbracket$ equivalent to a given word $w=uv\in\llbracket E\rrbracket^{I}$.

Definition 6.1**.**

The (unbounded and bounded) refined Antimirov $I$-reordering parts-of-derivatives of a regexp along a letter and a word are given by relations ${\to^{I}}\subseteq\mathsf{RE}\times\Sigma\times\mathsf{RE}\times\mathsf{RE}$, ${\Rightarrow^{I}}\subseteq\mathsf{RE}^{+}\times\Sigma\times\mathsf{RE}^{+}$, ${\to^{I*}}\subseteq\mathsf{RE}\times\Sigma^{*}\times\mathsf{RE}^{+}$, ${\Rightarrow^{I}_{N}}\subseteq\mathsf{RE}^{+\leq N+1}\times\Sigma\times\mathsf{RE}^{+\leq N+1}$, and ${\to^{I*}_{N}}\subseteq\mathsf{RE}\times\Sigma^{*}\times\mathsf{RE}^{+\leq N+1}$ defined inductively by

[TABLE]

[TABLE]

[TABLE]

By $\mathsf{RE}^{+\leq N+1}$ we mean nonempty lists of regexps of length at most $N+1$. The relations $\Rightarrow^{I}$ and $\to^{I*}$ are defined exactly as $\Rightarrow_{N}^{I}$ and $\to_{N}^{I*}$ but with the condition $|\Gamma,\Delta|<N$ of the first rule of $\Rightarrow_{N}^{I}$ dropped. The operation $R^{I}_{a}$ is extended to lists of regexps in the obvious way.

We have several rules for deriving a list of regexps along $a$. If $E$ is split into $E_{l},E_{r}$ and neither of them is nullable, then, in the $N$-bounded case, we require that the given list is shorter than $N+1$ since the new list will be longer by 1. If one of $E_{l},E_{r}$ is nullable, not the first resp. last in the list and we choose to drop it, then the new list will be of the same length. If both are nullable, not the first resp. last and we opt to drop both, then the new list will be shorter by 1. They must be droppable under these conditions to handle the situation when a word $z$ has been split as $v_{0}u_{1}v_{1}\ldots u_{k}v_{k}u_{k+1}\ldots u_{n}v_{n}$ and $v_{k}$ is further being split as $v_{l}av_{r}$ while $v_{l}$ or $v_{r}$ is empty. If $k\neq 0$ and $v_{l}$ is empty, we must join $u_{k}$ and $a$ into $u_{k}a$. If $k\neq n$ and $v_{r}$ is empty, we must join $a$ and $u_{k+1}$ into $au_{k+1}$. If $k$ is neither [math] nor $n$ and both $v_{l}$ and $v_{r}$ are empty, we must join all three of $u_{k}$, $a$ and $u_{k+1}$ into $u_{k}au_{k+1}$. The length of the new list of regexps is always at least 2.

Proposition 6.2**.**

For any $E$,

for any $a\in\Sigma,v_{l},v_{r}\in\Sigma^{*}$,

[TABLE] 2. 2.

for any $u\in\Sigma^{*},n\in\mathbb{N},v_{0}\in\Sigma^{*},v_{1},\ldots,v_{n-1}\in\Sigma^{+},v_{n}\in\Sigma^{*}$,

[TABLE]

Proof 6.3**.**

$\Longrightarrow$: By induction on $E$.

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Case $a^{\prime}$ where $a^{\prime}\neq a$: $v_{l}av_{r}\in\llbracket a^{\prime}\rrbracket=\{a^{\prime}\}$ is impossible.

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Case $a$: Suppose $v_{l}av_{r}\in\llbracket a\rrbracket=\{a\}$. Then $v_{l}=v_{r}=\varepsilon$. We have $a\to^{I}(a;1,1)$ and $\varepsilon\in\llbracket 1\rrbracket$ as required.

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Case [math]: $v_{l}av_{r}\in\llbracket 0\rrbracket=\emptyset$ is impossible.

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Case $E_{1}+E_{2}$: Suppose $v_{l}av_{r}\in\llbracket E_{1}+E_{2}\rrbracket=\llbracket E_{1}\rrbracket\cup\llbracket E_{2}\rrbracket$ and $v_{l}Ia$. Then $v_{l}av_{r}\in\llbracket E_{i}\rrbracket$ for one of two possible $i$. By IH for $E_{i},a,v_{l},v_{r}$, there are $E_{l}$, $E_{r}$ such that $E_{i}\to^{I}(a;E_{l},E_{r})$, $v_{l}\in\llbracket E_{l}\rrbracket$, $v_{r}\in\llbracket E_{r}\rrbracket$ and we also obtain $E_{1}+E_{2}\to^{I}(a;E_{l},E_{r})$.

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Case $1$: $v_{l}av_{r}\in\llbracket 1\rrbracket=\mathbf{1}$ is impossible.

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Case $EF$: Suppose $v_{l}av_{r}\in\llbracket EF\rrbracket=\llbracket E\rrbracket\cdot\llbracket F\rrbracket$ and $v_{l}Ia$. Then $v_{l}av_{r}=xy$ for some $x\in\llbracket E\rrbracket$ and $y\in\llbracket F\rrbracket$. Either (i) there exists $v^{\prime}$ such that $x=v_{l}av^{\prime}$ and $v_{r}=v^{\prime}y$ or (ii) there exists $v^{\prime}$ such that $y=v^{\prime}av_{r}$ and $v_{l}=xv^{\prime}$.

If (i), then, by IH for $E,a,v_{l},v^{\prime}$, there are $E_{l},E_{r}$ such that $E\to^{I}(a;E_{l},E_{r})$, $v_{l}\in\llbracket E_{l}\rrbracket$, $v^{\prime}\in\llbracket E_{r}\rrbracket$. We then also have $EF\to^{I}(a;E_{l},E_{r}F)$ and $v_{r}=v^{\prime}y\in\llbracket E_{r}F\rrbracket$.

If (ii), then $xIa$ and $v^{\prime}Ia$, so $x\in\llbracket R^{I}_{a}E\rrbracket$ and, by IH for $F,a,v^{\prime},v_{r}$, there are $F_{l},F_{r}$ such that $F\to^{I}(a;F_{l},F_{r})$, $v^{\prime}\in\llbracket F_{l}\rrbracket$, $v_{r}\in\llbracket F_{r}\rrbracket$. We then also have $EF\to^{I}(a;(R^{I}_{a}E)F_{l},F_{r})$ and $v_{l}=xv^{\prime}\in\llbracket(R^{I}_{a}E)F_{l}\rrbracket$.

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Case $E^{*}$: Suppose $v_{l}av_{r}\in\llbracket E^{*}\rrbracket$ and $v_{l}Ia$. Then $v_{l}=xv^{\prime}_{l}$, $v_{r}=v^{\prime}_{r}y$ for some $x,y\in\llbracket E^{*}\rrbracket$ and $v^{\prime}_{l},v^{\prime}_{r}$ such that $v^{\prime}_{l}av^{\prime}_{r}\in\llbracket E\rrbracket$. We have $xIa$, $v^{\prime}_{l}Ia$. Hence $x\in\llbracket R^{I}_{a}E^{*}\rrbracket$ and, by IH for $E,a,v^{\prime}_{l},v^{\prime}_{r}$, we get that there are $E_{l}$, $E_{r}$ such that $E\to^{I}(a;E_{l},E_{r})$ and $v^{\prime}_{l}\in\llbracket E_{l}\rrbracket$, $v^{\prime}_{r}\in\llbracket E_{r}\rrbracket$. We also obtain $v_{l}=xv^{\prime}_{l}\in\llbracket(R^{I}_{a}E^{*})E_{l}\rrbracket$ and $v_{r}=v^{\prime}_{r}y\in\llbracket E_{r}E^{*}\rrbracket$.

$\Longleftarrow$: By induction on the derivation of $E\to^{I}(a;E_{l},E_{r})$.

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Case $a\to^{I}(a;1,1)$ as an axiom: Suppose $v_{l},v_{r}\in\llbracket 1\rrbracket=\mathbf{1}$. Then $v_{l}=v_{r}=\varepsilon$ and we have $\varepsilon a\varepsilon=a\in\llbracket a\rrbracket$ as required.

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Case $E_{1}+E_{2}\to^{I}(a;E_{l},E_{r})$ inferred from $E_{i}\to^{I}(a;E_{l},E_{r})$ where $i$ is 1 or 2: Suppose $v_{l}\in\llbracket E_{l}\rrbracket$, $v_{r}\in\llbracket E_{r}\rrbracket$. We can then apply IH to the subderivation, $v_{l},v_{r}$ and obtain that $v_{l}Ia$ and $v_{l}av_{r}\in\llbracket E_{i}\rrbracket$, which gives us also that $v_{l}av_{r}\in\llbracket E_{1}\rrbracket\cup\llbracket E_{2}\rrbracket=\llbracket E_{1}+E_{2}\rrbracket$.

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Case $EF\to^{I}(a;E_{l},E_{r}F)$ inferred from $E\to^{I}(a;E_{l},E_{r})$: Suppose $v_{l}\in\llbracket E_{l}\rrbracket$, $v_{r}\in\llbracket E_{r}F\rrbracket=\llbracket E_{r}\rrbracket\cdot\llbracket F\rrbracket$. Then $v_{r}=xy$ for some $x\in\llbracket E_{r}\rrbracket$ and $y\in\llbracket F\rrbracket$. We can then apply IH to the subderivation, $v_{l},x$ and obtain that $v_{l}Ia$ and $v_{l}ax\in\llbracket E\rrbracket$. As $v_{r}=xy$, we obtain $v_{l}av_{r}=(v_{l}ax)y\in\llbracket E\rrbracket\cdot\llbracket F\rrbracket=\llbracket EF\rrbracket$.

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Case $EF\to^{I}(a,(R_{a}E)F_{l},F_{r})$ inferred from $F\to^{I}(a;F_{l},F_{r})$: Suppose $v_{l}\in\linebreak\llbracket(R_{a}E)F_{l}\rrbracket=R_{a}\llbracket E\rrbracket\cdot\llbracket F_{l}\rrbracket$, $v_{r}\in\llbracket F_{r}\rrbracket$. Then $v_{l}=xy$ for some $x\in R_{a}\llbracket E\rrbracket$ and $y\in\llbracket F_{l}\rrbracket$, which also gives us $xIa$ and $x\in\llbracket E\rrbracket$. We can then apply IH to the subderivation, $y,v_{r}$ and obtain that $yIa$ and $yav_{r}\in\llbracket F\rrbracket$. As $v_{l}=xy$, we get $v_{l}Ia$ and $v_{l}av_{r}=x(yav_{r})\in\llbracket E\rrbracket\cdot\llbracket F\rrbracket=\llbracket EF\rrbracket$.

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Case $E^{*}\to^{I}(a;(R_{a}E^{*})E_{l},E_{r}E^{*})$ inferred from $E\to^{I}(a;E_{l},E_{r})$: Suppose $v_{l}\in\llbracket(R_{a}E^{*})E_{l}\rrbracket=R_{a}\llbracket E^{*}\rrbracket\cdot\llbracket E_{l}\rrbracket$, $v_{r}\in\llbracket E_{r}E^{*}\rrbracket=\llbracket E_{r}\rrbracket\cdot\llbracket E^{*}\rrbracket$. Then $v_{l}=xy$ for some $x\in R_{a}\llbracket E^{*}\rrbracket$ and $y\in\llbracket E_{l}\rrbracket$, which also gives us $xIa$ and $x\in\llbracket E^{*}\rrbracket$, and $v_{r}=zw$ for some $z\in\llbracket E_{r}\rrbracket$ and $w\in\llbracket E^{*}\rrbracket$. We can then apply IH to the subderivation, $y,z$ and obtain that $yIa$ and $yaz\in\llbracket E\rrbracket$. As $v_{l}=xy$ and $v_{r}=zw$, we get that $v_{l}Ia$ and $v_{l}av_{r}=x(yaz)w\in\llbracket E^{*}\rrbracket\cdot\llbracket E\rrbracket\cdot\llbracket E^{*}\rrbracket\subseteq\llbracket E^{*}\rrbracket$. 2. 2.

$\Longrightarrow$: For any $E$ by induction on $u$.

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Case $\varepsilon$: Suppose $\varepsilon\sim^{I}u_{1}\ldots u_{n}$ and $z=v_{0}u_{1}v_{1}\ldots u_{n}v_{n}\in\llbracket E\rrbracket$. Then necessarily $n=0$, which means that we actually have $v_{0}\in\llbracket E\rrbracket$. We also have $E\to^{I^{*}}(\varepsilon;E)$ as required.

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Case $ua$: Suppose $ua\sim^{I}u_{1}\ldots u_{n}$ and $\forall i.\forall j<i.\,v_{j}Iu_{i}$ and $z=v_{0}u_{1}v_{1}\ldots u_{n}v_{n}\in\llbracket E\rrbracket$. It must be that $n>0$ and there must exist $k$ and $u_{l},u_{r}\in\Sigma^{*}$ such that $u_{k}=u_{l}au_{r}$, $v_{j}Ia$ for all $j<k$, $aIu_{r}$, $aIu_{i}$ for all $i>k$ and $u\sim^{I}u_{1}\ldots u_{k-1}u_{l}u_{r}u_{k+1}\ldots u_{n}$.

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If $u_{l}=u_{r}=\varepsilon$, then, as $v_{k-1}av_{k}Iu_{i}$ for all $i>k$, we are entitled to apply IH to $u$, $n-1$, $v_{0},v_{1},\ldots,v_{k-2},v_{k-1}av_{k},v_{k+1},\ldots,v_{n}$, $z$, $u_{1},\ldots,u_{k-1},u_{k+1},\ldots,u_{n}$. We get $E_{0},\ldots,E_{k-2}$, $E^{\prime}$, $E_{k+1}$, …$E_{n}$* such that $E\to^{I*}(u;E_{0},\ldots,E_{k-2},E^{\prime},E_{k+1},\ldots,E_{n})$ and $v_{j}\in\llbracket E_{j}\rrbracket$ for all $j<k-1$, $v_{k-1}av_{k}\in\llbracket E^{\prime}\rrbracket$, $v_{j}\in\llbracket E_{j}\rrbracket$ for all $j>k$. As $v_{k-1}Ia$, we can apply

1. to $E^{\prime}$, $a$, $v_{k-1}$, $v_{k}$ and get $E_{k-1},E_{k}$ such that $E^{\prime}\to^{I}(a;E_{k-1},E_{k})$, $v_{k-1}\in\llbracket E_{k-1}\rrbracket$ and $v_{k}\in\llbracket E_{k}\rrbracket$. This allows us to infer $E\to^{I*}(ua;R^{I}_{a}E_{0},\ldots,R^{I}_{a}E_{k-2},E_{k-1},E_{k},E_{k+1},\ldots,E_{n})$. As $v_{j}Ia$ also for all $j<k-1$, we in fact also have $v_{j}\in\llbracket R^{I}_{a}E_{j}\rrbracket$ for all $j<k-1$.*

• –

*If $u_{l}\neq\varepsilon$, $u_{r}=\varepsilon$, we note that $av_{k}Iu_{i}$ for all $i>k$ and apply IH to $u$, $n$, $v_{0}$, $v_{1},\ldots,v_{k-1}$, $av_{k},v_{k+1},\ldots,v_{n}$, $z$, $u_{1},\ldots,u_{k-1},u_{l},u_{k+1},\ldots,u_{n}$. We get $E_{0},\ldots,E_{k-1}$, $E^{\prime}$, $E_{k+1},\ldots E_{n}$ such that $E\to^{I*}(u;E_{0},\ldots,E_{k-1},E^{\prime},E_{k+1},\ldots,E_{n})$ and $v_{j}\in\llbracket E_{j}\rrbracket$ for all $j<k$, $av_{k}\in\llbracket E^{\prime}\rrbracket$, $v_{j}\in\llbracket E_{j}\rrbracket$ for all $j>k$. As $\varepsilon Ia$, we can apply

1. to $E^{\prime}$, $a$, $\varepsilon$, $v_{k}$ and get $E^{\prime\prime},E_{k}$ such that $E^{\prime}\to^{I}(a;E^{\prime\prime},E_{k})$, $\varepsilon\in\llbracket E^{\prime\prime}\rrbracket$, $v_{k}\in\llbracket E_{k}\rrbracket$. As $\varepsilon\in\llbracket E^{\prime\prime}\rrbracket$ tells us that $E^{\prime\prime}{\downarrow}$, we can infer $E\to^{I*}(ua;R^{I}_{a}E_{0},\ldots,R^{I}_{a}E_{k-1},E_{k},E_{k+1},\ldots,E_{n})$. As $v_{j}Ia$ also for all $j<k$, we in fact also have $v_{j}\in\llbracket R^{I}_{a}E_{j}\rrbracket$ for all $j<k$.*

• –

The cases $u_{l}=\varepsilon$, $u_{r}\neq\varepsilon$ and $u_{l}\neq\varepsilon$, $u_{r}\neq\varepsilon$ are handled similarly to the previous case.

$\Longleftarrow$: By induction on the derivation of $E\to^{I*}(u;E_{0},\ldots,E_{n})$.

• •

Case $E\to^{I*}(\varepsilon;E)$ as an axiom: Suppose that $v_{0}\in\llbracket E\rrbracket$. We have $\varepsilon\sim\varepsilon$ as well as $z=v_{0}\in\llbracket E\rrbracket$ directly.

• •

Case $E\to^{I*}(ua;R^{I}_{a}E_{0},\ldots,R^{I}_{a}E_{k-1},E_{l},E_{r},E_{k+1},\ldots,E_{n})$ inferred from subderivations $E\to^{I*}(u;E_{0},\ldots,E_{n})$ and $E_{k}\to^{I}(a;E_{l},E_{r})$: Suppose that $v_{0}\in\llbracket R^{I}_{a}E_{0}\rrbracket$, …, $v_{k-1}\in\llbracket R^{I}_{a}E_{k-1}\rrbracket$, $v_{l}\in\llbracket E_{l}\rrbracket$, $v_{r}\in\llbracket E_{r}\rrbracket$, $v_{k+1}\in\llbracket E_{k+1}\rrbracket$, …, $v_{n}\in\llbracket E_{n}\rrbracket$, which gives us $v_{0}Ia$, …, $v_{k-1}Ia$, $v_{0}\in\llbracket E_{0}\rrbracket$, …, $v_{k-1}\in\llbracket E_{k-1}\rrbracket$. Applying (1.$\Longleftarrow$) to $E_{k}$, $a$, $v_{l}$, $v_{r}$, we learn that $v_{l}Ia$ and $v_{l}av_{r}\in\llbracket E_{k}\rrbracket$. Applying IH to the subderivation, $v_{0},\ldots,v_{k-1}$, $v_{l}av_{r}$, $v_{k+1},\ldots,v_{n}$, we obtain $z,u_{1},\ldots,u_{n}\in\Sigma^{+}$ such that $u\sim^{I}u_{1}\ldots u_{n}$, $\forall i.\,\forall j<i.\ v_{j}Iu_{i}$, $\forall i>k.\,v_{l}av_{r}Iu_{i}$ and $z=v_{0}u_{1}v_{1}\ldots v_{k-1}u_{k}(v_{l}av_{r})u_{k+1}v_{k+1}\ldots u_{n}v_{n}\in\llbracket E\rrbracket$. Clearly $ua\sim^{I}u_{1}\ldots u_{k}au_{k+1}\ldots u_{n}$ and $z=v_{0}u_{1}v_{1}\ldots v_{k-1}u_{k}v_{l}av_{r}u_{k+1}v_{k+1}\ldots u_{n}v_{n}\in\llbracket E\rrbracket$.

• •

Case $E\to^{I*}(ua;R^{I}_{a}E_{0},\ldots,R^{I}_{a}E_{k-1},E_{r},E_{k+1},\ldots,E_{n})$ inferred from subderivations $E\to^{I*}(u;E_{0},\ldots,E_{n})$ and $E_{k}\to^{I}(a;E_{l},E_{r})$ and $E_{l}{\downarrow}$ whereby $k\neq 0$: Suppose that $v_{0}\in\llbracket R^{I}_{a}E_{0}\rrbracket$, …, $v_{k-1}\in\llbracket R^{I}_{a}E_{k-1}\rrbracket$, $v_{r}\in\llbracket E_{r}\rrbracket$, $v_{k+1}\in\llbracket E_{k+1}\rrbracket$, …, $v_{n}\in\llbracket E_{n}\rrbracket$, which gives us $v_{0}Ia$, …, $v_{k-1}Ia$, $v_{0}\in\llbracket E_{0}\rrbracket$, …, $v_{k-1}\in\llbracket E_{k-1}\rrbracket$. Applying (1.$\Longleftarrow$) to $E$, $a$, $\varepsilon$, $v_{r}$, we learn that $av_{r}\in\llbracket E_{k}\rrbracket$. Applying IH to the subderivation, $v_{0},\ldots,v_{k-1}$, $av_{r}$, $v_{k+1},\ldots,v_{n}$, we obtain $z,u_{1},\ldots,u_{n}\in\Sigma^{+}$ such that $u\sim^{I}u_{1}\ldots u_{n}$, $\forall i.\,\forall j<i.\ v_{j}Iu_{i}$, $\forall i>k.\,av_{r}Iu_{i}$ and $z=v_{0}u_{1}v_{1}\ldots v_{k-1}u_{k}(av_{r})u_{k+1}v_{k+1}\ldots u_{n}v_{n}\in\llbracket E\rrbracket$. Now clearly $ua\sim^{I}u_{1}\ldots(u_{k}a)u_{k+1}\ldots u_{n}$ and $z=v_{0}u_{1}v_{1}\ldots v_{k-1}(u_{k}a)v_{r}u_{k+1}v_{k+1}\ldots u_{n}v_{n}\in\llbracket E\rrbracket$.

• •

The two remaining cases are treated similarly to the previous case.

Proposition 6.4**.**

For any $E$,

for any $a\in\Sigma,v\in\Sigma^{*}$, the following are equivalent:

1. (a)

$av\in\llbracket E\rrbracket^{I}$; 2. (b)

$\exists v_{l},v_{r}\in\Sigma^{*}.\,v\sim^{I}v_{l}v_{r}\wedge v_{l}Ia\wedge v_{l}av_{r}\in\llbracket E\rrbracket$; 3. (c)

$\exists v_{l},v_{r}\in\Sigma^{*}.\,\\ \hskip 14.22636ptv\sim^{I}v_{l}v_{r}\wedge\exists E_{l},E_{r}.\,E\to^{I}(a;E_{l},E_{r})\wedge v_{l}\in\llbracket E_{l}\rrbracket\wedge v_{r}\in\llbracket E_{r}\rrbracket$; 4. (d)

$\exists v_{l},v_{r}\in\Sigma^{*}.\,\\ \hskip 14.22636ptv\in v_{l}\cdot^{I}v_{r}\wedge\exists E_{l},E_{r}.\,E\to^{I}(a;E_{l},E_{r})\wedge v_{l}\in\llbracket E_{l}\rrbracket^{I}\wedge v_{r}\in\llbracket E_{r}\rrbracket^{I}$. 2. 2.

for any $u,v\in\Sigma^{*}$, the following are equivalent:

1. (a)

$uv\in\llbracket E\rrbracket^{I}$; 2. (b)

$\exists z\in\llbracket E\rrbracket.\ u\mathrel{{\sim}{\lhd}}z\mathrel{{\rhd}{\sim}}v$; 3. (c)

$\exists n\in\mathbb{N},v_{0}\in\Sigma^{*},v_{1},\ldots,v_{n-1}\in\Sigma^{+},v_{n}\in\Sigma^{*}.\,v\sim^{I}v_{0}v_{1}\ldots v_{n}\wedge\\ \hskip 14.22636pt\exists E_{0},\ldots,E_{n}.\,E\to^{I*}(u;E_{0},\ldots,E_{n})\wedge\forall j.\,v_{j}\in\llbracket E_{j}\rrbracket$; 4. (d)

$\exists n\in\mathbb{N},v_{0}\in\Sigma^{*},v_{1},\ldots,v_{n-1}\in\Sigma^{+},v_{n}\in\Sigma^{*}.\,v\in v_{0}\cdot^{I}v_{1}\cdot^{I}\ldots\cdot^{I}v_{n}\wedge\\ \hskip 14.22636pt\exists E_{0},\ldots,E_{n}.\,E\to^{I*}(u;E_{0},\ldots,E_{n})\wedge\forall j.\,v_{j}\in\llbracket E_{j}\rrbracket^{I}$. 3. 3.

*for any $u\in\Sigma^{*}$, *

*  $u\in\llbracket E\rrbracket^{I}\iff(u=\varepsilon\wedge E{\downarrow})\vee(u\neq\varepsilon\wedge\exists E_{0},E_{1}.\,E\to^{I*}(u;E_{0},E_{1})\wedge E_{0}{\downarrow}\wedge E_{1}{\downarrow})$.*

Proof 6.5**.**

(a) $\iff$ (b) follows from Proposition 3.6. (b) $\iff$ (c) follows from Proposition 6.2(1). (c) $\iff$ (d) follows from Proposition 3.10. 2. 2.

(a) $\iff$ (b) follows from Proposition 3.6. (b) $\iff$ (c) follows from Proposition 6.2(2). (c) $\iff$ (d) follows from Proposition 3.10. 3. 3.

From (2) for $E,u,\varepsilon$.

Proposition 6.6**.**

For any $E$, $N\in\mathbb{N}$, $u\in\Sigma^{*}$, $z\in\llbracket E\rrbracket$,

[TABLE]

Proof 6.7**.**

By replaying the proof of Proposition 6.2(2.$\Longrightarrow$). In the fourth subcase ($u_{l}\neq\varepsilon$, $u_{r}\neq\varepsilon$) of the case $ua$ of induction, IH for $u^{\prime}=u$, $u^{\prime\prime}=a$ is needed.

Corollary 6.8**.**

For any $E$ such that $\llbracket E\rrbracket$ has uniform rank at most $N$,

for any $u,v\in\Sigma^{*}$, the following are equivalent:

1. (a)

$uv\in\llbracket E\rrbracket^{I}$; 2. (b)

$\exists z\in\llbracket E\rrbracket.\ \forall u^{\prime},u^{\prime\prime}.\,u=u^{\prime}u^{\prime\prime}\Longrightarrow u^{\prime}\mathrel{{\sim}{\lhd}}_{N}z\mathrel{{\rhd}{\sim}}u^{\prime\prime}v$; 3. (c)

$\exists n\leq N,v_{0}\in\Sigma^{*},v_{1},\ldots,v_{n-1}\in\Sigma^{+},v_{n}\in\Sigma^{*}.\,v\sim^{I}v_{0}v_{1}\ldots v_{n}\wedge\\ \hskip 14.22636pt\exists E_{0},\ldots,E_{n}.\,E\to^{I*}_{N}(u;E_{0},\ldots,E_{n})\wedge\forall j.\,v_{j}\in\llbracket E_{j}\rrbracket$; 4. (d)

$\exists n\leq N,v_{0}\in\Sigma^{*},v_{1},\ldots,v_{n-1}\in\Sigma^{+},v_{n}\in\Sigma^{*}.\,v\in v_{0}\cdot^{I}v_{1}\cdot^{I}\ldots\cdot^{I}v_{n}\wedge\\ \hskip 14.22636pt\exists E_{0},\ldots,E_{n}.\,E\to^{I*}_{N}(u;E_{0},\ldots,E_{n})\wedge\forall j.\,v_{j}\in\llbracket E_{j}\rrbracket^{I}$. 2. 2.

*for any $u\in\Sigma^{*}$, *

*  $u\in\llbracket E\rrbracket^{I}\iff(u=\varepsilon\wedge E{\downarrow})\vee(u\neq\varepsilon\wedge\exists E_{0},E_{1}.\,E\to^{I*}_{N}(u;E_{0},E_{1})\wedge E_{0}{\downarrow}\wedge E_{1}{\downarrow})$.*

Proof 6.9** (Proof of 1).**

(a) $\Longrightarrow$ (b) is from $E$ having uniform rank at most $N$. (b) $\Longrightarrow$ (c) follows from Proposition 6.6. (c) $\Longrightarrow$ (d) and (d) $\Longrightarrow$ (a) are those from Proposition 6.4.

Example 6.10**.**

We go back to Example 4.11. Recall that $E=_{\mathrm{df}}aa+ab+b$ and $E_{b}=_{\mathrm{df}}R^{I}_{b}E=aa+a0+0$. Here is one of the refined reordering parts-of-derivatives of $E^{*}$ along $bb$.

[TABLE]

In this example, we chose $N=_{\mathrm{df}}2$. The regexp $1(E_{b}^{*}(a1))\doteq(aa)^{*}a$ is not nullable, so we could not have dropped it. From here we cannot continue by deriving along a third $b$ by again taking it from the summand $ab$ of $E$ in $1E^{*}$, as this would produce another nondroppable $1(E_{b}^{*}(a1))$ and make the list too long (longer than 3). For example, we are not allowed to establish $w=_{\mathrm{df}}bbbaaa\in\llbracket E^{*}\rrbracket^{I}$ (by deriving $E^{*}$ along $w$ and checking if we can arrive at $E_{0},E_{1}$ with both $E_{0},E_{1}$ nullable), mandated by $z=_{\mathrm{df}}ababab\in\llbracket E^{*}\rrbracket$, but we are allowed to do so because of $z^{\prime}=_{\mathrm{df}}bbabaa\in\llbracket E^{*}\rrbracket$. The word $z$ is not useful since among the splits of $w$ as $w=uv$ there is $u=_{\mathrm{df}}bbb$, $v=_{\mathrm{df}}aaa$, which splits $z$ as $u\mathrel{{\sim}{\lhd}}z\mathrel{{\rhd}{\sim}}v$ scattering $u$ into 3 blocks as $z=a\underline{b}a\underline{b}a\underline{b}$ (we underline the letters from $u$); the full sequence of these corresponding splits of $z$ is $ababab$, $a\underline{b}abab$, $a\underline{b}a\underline{b}ab$, $a\underline{b}a\underline{b}a\underline{b}$, $\underline{ab}a\underline{b}a\underline{b}$, $\underline{abab}a\underline{b}$, $\underline{ababab}$. The word $z^{\prime}$, on the contrary, is fine because, for every split of $w$ as $w=uv$, there are at most two blocks of letters from $u$ in $z^{\prime}$: $bbabaa$, $\underline{b}babaa$, $\underline{bb}abaa$, $\underline{bb}a\underline{b}aa$, $\underline{bbab}aa$, $\underline{bbaba}a$, $\underline{bbabaa}$. The choice $N=2$ suffices for accepting all of $\llbracket E^{*}\rrbracket^{I}$, since $\llbracket E^{*}\rrbracket$ happens to have uniform rank 2.

The refined Antimirov reordering parts-of-derivatives of a regexp $E$ give a nondeterministic automaton by $Q^{E}=_{\mathrm{df}}\{\Gamma\mid\exists u\in\Sigma^{*}.\,E\to^{I*}(u;\Gamma)\}$, $I^{E}=_{\mathrm{df}}\{E\}$, $F^{E}=_{\mathrm{df}}\{E\mid E{\downarrow}\}\cup\{E_{0},E_{1}\in Q^{E}\mid E_{0}{\downarrow}\wedge E_{1}{\downarrow}\}$, $\Gamma\to^{E}(a;\Gamma^{\prime})=_{\mathrm{df}}\Gamma\Rightarrow^{I}(a;\Gamma^{\prime})$ for $\Gamma,\Gamma^{\prime}\in Q^{E}$. By Prop. 6.4, this automaton accepts $\llbracket E\rrbracket^{I}$. It is generally not finite as $Q^{E}$ can contain states $\Gamma$ of any length.

Given $N\in\mathbb{N}$, another automaton is obtained by restricting $Q^{E}$, $F^{E}$ and $\to^{E}$ to $Q^{E}_{N}=_{\mathrm{df}}\{\Gamma\mid\exists u\in\Sigma^{*}.\,E\to^{I*}_{N}(u;\Gamma)\}$, $F^{E}_{N}=_{\mathrm{df}}\{E\mid E{\downarrow}\}\cup\{E_{0},E_{1}\in Q^{E}_{N}\mid E_{0}{\downarrow}\wedge E_{1}{\downarrow}\}$, $\Gamma\to^{E}_{N}(a;\Gamma^{\prime})=_{\mathrm{df}}\Gamma\Rightarrow^{I}_{N}(a;\Gamma^{\prime})$ for $\Gamma,\Gamma^{\prime}\in Q^{E}_{N}$. By Cor. 6.8, if $\llbracket E\rrbracket$ has uniform rank at most $N$, then this smaller automaton accepts $\llbracket E\rrbracket^{I}$ despite the truncation. If $\llbracket E\rrbracket$ does not have uniform rank or we choose $N$ smaller than the uniform rank, then the $N$-truncated automaton recognizes a proper subset of $\llbracket E\rrbracket^{I}$. Prop. 5.7 gives an example of this: however we choose $N$, the $N$-truncated automaton fails to accept the word $a^{n}b^{n}ca^{n}b^{n}$ for $n>N$. This happens because $\llbracket E\rrbracket$ does not have uniform rank (and that it has rank 2 does not help).

6.2 Automaton Finiteness for Regular Expressions with Uniform Rank

Is the $N$-truncated Antimirov automaton finite? The states $\Gamma$ of $Q^{E}_{N}$ are all of length at most $N+1$, so there is hope. The automaton will be finite if we can find a finite set containing all the individual regexps $E^{\prime}$ appearing in the states $\Gamma$. We now define such a set $E^{\to*}$.

Definition 6.11**.**

We define functions $(\_)^{\rightsquigarrow+},\mathbf{R},(\_)^{\to+},(\_)^{\to*}:\mathsf{RE}\to\mathcal{P}\mathsf{RE}$ by

[TABLE]

[TABLE]

Proposition 6.12**.**

For any $E$, the set $E^{\to*}$ is finite. 2. 2.

For any $E$ and $X$, we have $(R^{I}_{X}E)^{\to*}\subseteq R^{I}_{X}(E^{\to*})$. 3. 3.

For any $E$, $a$ and $E_{l},E_{r}$, if $E\to^{I}(a;E_{l},E_{r})$, then $E_{l}\in R^{I}_{a}(E^{\rightsquigarrow+})$ and $E_{r}\in E^{\rightsquigarrow+}$. 4. 4.

*For any $E,E^{\prime},X,a,E^{\prime}_{l},E^{\prime}_{r}$, if $E^{\prime}\in R^{I}_{X}(E^{\rightsquigarrow+})$ and $E^{\prime}\to^{I}(a;E^{\prime}_{l},E^{\prime}_{r})$, *

then $E^{\prime}_{l}\in R^{I}_{Xa}(E^{\rightsquigarrow+})$ and $E^{\prime}_{r}\in R^{I}_{X}(E^{\rightsquigarrow+})$. 5. 5.

*For any $E$, $u$ and $E_{0},\ldots,E_{n}$, if $E\to^{I*}(u;E_{0},\ldots,E_{n})$, then $\forall j.\,E_{j}\in E^{\to*}$. *

Proposition 6.13**.**

For every $E$ and $N$, the state set $\{\Gamma\mid\exists u\in\Sigma^{*}.\,E\to^{I*}_{N}(u;\Gamma)\}$ of the $N$-truncated refined Antimirov automaton for $E$ (accepting $\llbracket E\rrbracket^{I}$ if $\llbracket E\rrbracket$ has uniform rank at most $N$) is finite.

7 Related Work

Syntactic derivative constructions for regular expressions extended with constructors for (versions of) the shuffle operation have been considered, for example, by Sulzmann and Thiemann [23] for the Brzozowski derivative and by Broda et al. [6] for the Antimirov derivative. This is relevant to our derivatives since $L\cdot^{I}L^{\prime}$ is by definition a language between $L\cdot L^{\prime}$ and $L\mathrel{{\sqcup}{\sqcup}}L^{\prime}$. Thus our Brzozowski and Antimirov reordering derivatives of $EF$ must be between the classical Brzozowski and Antimirov derivatives of $EF$ and $E\mathrel{{\sqcup}{\sqcup}}F$.

8 Conclusion and Future Work

We have shown that the Brzozowski and Antimirov derivative operations generalize to trace closures of regular languages in the form of reordering derivative operations. The sets of Brzozowski resp. Antimirov reordering (parts-of-)derivatives of a regexp are generally infinite, so the deterministic and nondeterministic automata that they give, accepting the trace closure, are generally infinite. Still, if the regexp is star-connected, their appropriate quotients are finite. Also, the set of $N$-bounded refined Antimirov reordering parts-of-derivatives is finite without quotienting, and we showed that, if the language of the regexp has uniform rank at most $N$, the $N$-truncated refined Antimirov automaton accepts the trace closure. We also proved that star-connected expressions define languages with finite uniform rank.

In summary, we have established the following picture.

[TABLE]

Our intended application of reordering derivatives is operational semantics in the context of relaxed memory (where, e.g., shadow writes, i.e., writes from local buffers to shared memory, can be reorderable with other actions). For sequential composition $EF$ it is usually required that, to execute any action from $F$, execution of $E$ must have completed. In the jargon of derivatives, this is to say that for an action from $F$ to become executable, what is left of $E$ has to have become nullable (i.e., one can consider the execution of $E$ completed). With reordering derivatives, we can execute an action from $F$ successfully even when what is left of $E$ is not nullable. It suffices that some sequence of actions to complete the residual of $E$ is reorderable with the selected action of $F$.

In the definitions of the derivative operations we only use $I$ in one direction, i.e., we do not make use of its symmetry. It would be interesting to see if our results can be generalized to the setting of semi-commutations [8] and which changes are required for that.